Unit 20 · Geometry: Reasoning, Congruence and Triangles

Picture a map of Chicago. A bus stop is a dot on the map. Lake Shore Drive is a long straight stroke. The flat surface of Lake Michigan spreads out in every direction. In geometry these three pictures stand for the three undefined terms: point, line and plane. We describe them, but we never define them using other words, because every chain of definitions has to start somewhere. A point has no size, only a position. A line is perfectly straight, has no thickness, and continues forever in both directions. A plane is a flat surface with no edges.

Everything else gets a definition built from those three. A segment is the part of a line between two endpoints, like the stretch of Michigan Avenue between two cross streets. A ray starts at one endpoint and continues forever in one direction. Points that lie on the same line are collinear. Points that lie on the same plane are coplanar. When two different lines cross, they share exactly one point, called their intersection. Read each definition out loud the first time you meet it. A definition is a promise about what a word will mean every single time you use it.

Names matter. The line through points A and B is called line AB. Segment AB has endpoints A and B. In plain text, the length of that segment is also written AB, so you have to read carefully. Here is a common mistake: treating segment AB and the length AB as the same thing. One is a set of points you can draw. The other is a number, like 7 cm. Keep the object and its measurement separate in your head. Proofs later in this chapter become much easier when you know whether you are talking about a thing or a number.

Every game needs rules that players agree on before play begins. Geometry’s starting rules are called postulates, and each one is a postulate. Some books call them axioms. A postulate is a statement accepted as true without proof. Euclid used five. One says that through any two points there is exactly one line. Another says a segment can be extended into a line. You cannot prove these from anything simpler. You accept them, and then you build.

A theorem is a statement that has been proved. The proof uses only definitions, postulates and theorems that were already proved. Here is a small one: if two different lines intersect, they intersect in exactly one point. Suppose two different lines shared two points, P and Q. Then both lines would pass through P and Q. But the postulate says exactly one line passes through two given points, so the two lines would have to be the same line. That contradicts the fact that they were different. So they can share at most one point. Notice how the proof leans on the postulate at the key moment.

Some postulates are about measuring. The segment addition postulate says that if point B is between A and C, then AB + BC = AC. Try it with algebra. Suppose AB = 3x, BC = 2x + 4 and AC = 29. Then 3x + 2x + 4 = 29, so 5x = 25 and x = 5. That gives AB = 15 and BC = 14. Check: 15 + 14 = 29. The angle addition postulate works the same way for two angles that share a side. A common mistake is to solve for x and stop. Read the question again. It may ask for AB, not for x.

Stand at a bus stop every morning and watch. The bus comes at 7:05, then 7:20, then 7:35, then 7:50. You guess the next one arrives at 8:05. That is inductive reasoning: looking at several cases, noticing a pattern, and making a conjecture, which is an educated guess that might be true. Inductive reasoning is how most math ideas are born. But it is not proof. The bus schedule might change at 8:00.

Deductive reasoning starts from facts already accepted and reaches a conclusion that must follow. All right angles measure 90°. Angle X is a right angle. So angle X measures 90°. Here is a number example. Conjecture: the sum of two odd numbers is even. Inductive check: 3 + 5 = 8, 7 + 9 = 16, 11 + 1 = 12. Deductive proof: any odd number can be written 2m + 1, and another as 2n + 1. Their sum is 2m + 2n + 2 = 2(m + n + 1), which is 2 times a whole number, so it is even. The proof covers every odd pair at once, not just three of them.

A counterexample is one case where the pattern breaks. One is enough to sink a conjecture. Look at the formula n^2 + n + 41. For n = 1 it gives 43, for n = 2 it gives 47, for n = 3 it gives 53, for n = 4 it gives 61. All prime. It keeps producing primes for a long time, so a student might conjecture it always does. But when n = 40, the formula gives 1600 + 40 + 41 = 1681, and 41 × 41 = 1681. Not prime. Two common mistakes: believing that many examples prove a rule, and believing that one example proves it. Examples suggest. Proofs settle.

If it rains on Saturday, then the game is canceled. That sentence is a conditional statement. The part after if is the hypothesis: it rains on Saturday. The part after then is the conclusion: the game is canceled. Mathematicians write a conditional as p → q, read as if p then q. Many geometry facts hide their if-then shape. Vertical angles are congruent really means: if two angles are vertical angles, then they are congruent. Rewriting a fact in if-then form shows you exactly what you are assuming and exactly what you are claiming.

When is a conditional false? Only in one situation: the hypothesis happens and the conclusion does not. It rains, and the game is played anyway. That breaks the promise. If it does not rain, the promise was never tested, so the statement is still counted as true. Try a number example. If x = 3, then x^2 = 9. That is true, since 3 × 3 = 9. Now flip it: if x^2 = 9, then x = 3. That is false, because x could be −3, and (−3) × (−3) = 9 too. The first statement makes a promise it keeps. The second makes a promise it breaks.

Two common mistakes. First, assuming the hypothesis always comes first in the sentence. You pass the class if you earn 60 percent puts the hypothesis at the end: earning 60 percent is the if part. Second, treating a conditional as if it works both ways. A conditional only promises one direction. The negation of a statement, written not p, says the opposite. The negation of x is greater than 5 is x is not greater than 5, which includes x = 5. Watch that edge case.

Start with a true statement: if a number ends in 0, then it is divisible by 5. Now build three relatives. The converse swaps the two parts: if a number is divisible by 5, then it ends in 0. False, because 15 is divisible by 5 and ends in 5. The inverse negates both parts: if a number does not end in 0, then it is not divisible by 5. Also false, and 15 is again the counterexample. The contrapositive swaps and negates: if a number is not divisible by 5, then it does not end in 0. True. Every number ending in 0 is divisible by 5, so a number that is not divisible by 5 cannot end in 0.

That pattern is not a coincidence. A conditional and its contrapositive are logically equivalent: they are always both true or both false. Think about the rain promise. If it rains, the game is canceled. Suppose the game is not canceled. Then it must not have rained, or the promise would have been broken. So, if the game is not canceled, then it did not rain. Same promise, stated backward. The converse and inverse are also equivalent to each other, but neither is tied to the original.

In geometry this matters because converses often need their own proofs, and some converses are false. If two angles are vertical angles, then they are congruent: true. The converse, if two angles are congruent then they are vertical angles, is false, since two separate 30° angles are congruent without being vertical. When a proof seems stuck, try proving the contrapositive instead. Showing that not q leads to not p proves the original just as well. A common mistake is to assume a true statement’s converse is automatically true. Check it separately, every time.

Two lines are perpendicular lines if and only if they meet at a right angle. The phrase if and only if means the statement runs both ways. If the lines are perpendicular, they meet at a right angle. And if they meet at a right angle, they are perpendicular. A statement of this shape is a biconditional, written p ↔ q. It is true only when the conditional and its converse are both true. If either direction fails, the biconditional is false.

Every good definition is a biconditional, even when it is not written that way. A right angle is an angle that measures 90°. Read forward: if an angle is a right angle, it measures 90°. Read backward: if an angle measures 90°, it is a right angle. Both hold, so the definition works. Now test a bad one: a square is a quadrilateral with four right angles. Forward is fine, since every square has four right angles. Backward fails: a 3 by 5 rectangle has four right angles but is not a square. So that sentence is not a definition of a square. It describes a rectangle.

Here is an algebra biconditional: x + 5 = 12 if and only if x = 7. Forward: subtract 5 from both sides, and x = 7. Backward: 7 + 5 = 12. Both directions work, so the biconditional is true. Compare it with x^2 = 49 if and only if x = 7. Backward is fine, since 7^2 = 49. Forward fails, because x = −7 also gives 49. That biconditional is false. The common mistake is writing if and only if when only one direction has been checked. Always test both.

You already write proofs when you solve equations; you just do not label the reasons. Solve 3x + 7 = 22. Statement 1: 3x + 7 = 22. Reason: given. Statement 2: 3x = 15. Reason: subtraction property of equality, since subtracting 7 from both sides keeps the equation true. Statement 3: x = 5. Reason: division property of equality. That is a two-column proof. Every line on the left has a reason on the right, and each reason is a definition, a postulate, a property or a theorem. Other properties you will use: the reflexive property (a = a), the symmetric property (if a = b then b = a), the transitive property (if a = b and b = c then a = c) and the substitution property (if a = b, you may replace a with b anywhere).

Now a segment proof. Points A, B, C and D lie on a line in that order. Given: AB = CD. Prove: AC = BD. Line 1: AB = CD, given. Line 2: BC = BC, reflexive property. Line 3: AB + BC = CD + BC, addition property of equality, since we added the same length to both sides. Line 4: AB + BC = AC and CD + BC = BD, segment addition postulate. Line 5: AC = BD, substitution. Check it with numbers: put A at 0, B at 2, C at 5 and D at 7. Then AB = 2 and CD = 2, while AC = 5 and BD = 5. The proof says it works for any positions, not just these.

Three habits keep proofs honest. Every statement needs a reason, even the obvious ones. The word given may only be used for facts the problem handed you. And you may never use the statement you are trying to prove as a reason along the way, because that is circular. A common mistake is skipping the reflexive line, BC = BC. It looks silly, but without it there is nothing to justify adding BC to both sides.

Two streets cross, and four angles appear at the corner. Two angles that share a side and whose other sides form a straight line are a linear pair. The linear pair postulate says their measures add to 180°, so they are supplementary. Angles that add to 90° are complementary. The two angles across from each other at the crossing are vertical angles. Number the four angles 1, 2, 3, 4 going around. Angles 1 and 3 are vertical, and so are 2 and 4.

Here is the proof that vertical angles are congruent, and it uses only the linear pair postulate. Angles 1 and 2 form a linear pair, so m∠1 + m∠2 = 180°. Angles 3 and 2 also form a linear pair, so m∠3 + m∠2 = 180°. Both sums equal 180°, so m∠1 + m∠2 = m∠3 + m∠2 by substitution. Subtract m∠2 from both sides: m∠1 = m∠3. That is the subtraction property of equality. Try it with numbers. If vertical angles measure (3x + 10)° and (5x − 30)°, set them equal: 3x + 10 = 5x − 30, so 40 = 2x and x = 20. Each angle is 70°. Its linear pair partner is 180° − 70° = 110°.

The same subtraction trick proves two more theorems. If two angles are supplementary to the same angle, they are congruent, because each one equals 180° minus that angle. If two angles are complementary to the same angle, they are congruent for the same reason with 90°. Two common mistakes: setting vertical angles equal to 180° instead of equal to each other, and assuming two angles are congruent because they look alike in a drawing. In a proof, a picture is a hint, not a reason.

A midpoint is the point that divides a segment into two congruent segments. Claim: if M is the midpoint of AB, then AM = 1/2 AB. Here it is as a paragraph proof, which is a proof written in sentences. Since M is the midpoint of AB, AM = MB by the definition of midpoint. By the segment addition postulate, AM + MB = AB. Substituting AM for MB gives AM + AM = AB, so 2 AM = AB. Dividing both sides by 2 gives AM = 1/2 AB. Every sentence names its reason, exactly as a two-column proof would.

A flow proof shows the same logic as boxes connected by arrows, so you can see which facts feed into which. One box holds M is the midpoint of AB (given). An arrow leads to AM = MB (definition of midpoint). A second starting box holds AM + MB = AB (segment addition postulate). Both arrows meet at 2 AM = AB (substitution), which leads to AM = 1/2 AB (division property). Flow proofs are useful when two separate chains of reasoning join at one conclusion, because the arrows show the joining.

Check the result with numbers. If AB = 18, then AM should be 9, and 9 + 9 = 18. Now use it in algebra: M is the midpoint of AB, with AM = 3x + 2 and MB = 5x − 8. Since AM = MB, 3x + 2 = 5x − 8, so 10 = 2x and x = 5. Then AM = 17, MB = 17 and AB = 34. The most common mistake in paragraph proofs is dropping the reasons because the sentences flow smoothly. A paragraph proof is still a proof. Every claim has to say why.

Chicago’s streets mostly run in a grid, with east-west streets parallel to each other. A few diagonal streets, like Milwaukee Avenue, slice across the grid. A line that crosses two or more other lines is a transversal. Where a transversal crosses two lines, eight angles appear, four at each crossing. Number them 1 through 4 at the top crossing and 5 through 8 at the bottom, in the same order. Angles in the same position at each crossing, like 1 and 5, are corresponding angles. Angles between the two lines on opposite sides of the transversal, like 3 and 6, are alternate interior angles. Angles between the lines on the same side, like 3 and 5, are consecutive interior angles.

The corresponding angles postulate says: if the two lines are parallel lines, then corresponding angles are congruent. From that one rule we can prove the rest. Claim: alternate interior angles 3 and 6 are congruent. Angle 2 corresponds to angle 6, so ∠2 ≅ ∠6. Angle 2 and angle 3 are vertical angles, so ∠2 ≅ ∠3. By the transitive property, ∠3 ≅ ∠6. Consecutive interior angles are supplementary by a similar chain: ∠3 ≅ ∠6 as just shown, and angles 6 and 5 form a linear pair, so m∠3 + m∠5 = m∠6 + m∠5 = 180°.

Now some numbers. Alternate interior angles measure (2x + 15)° and (3x − 20)°. They are congruent, so 2x + 15 = 3x − 20 and x = 35. Each angle is 85°. A consecutive interior angle next to one of them measures 180° − 85° = 95°. The converses also hold and are how we prove lines parallel: if corresponding angles are congruent, or alternate interior angles are congruent, or consecutive interior angles are supplementary, then the lines are parallel. Two mistakes to avoid: assuming lines are parallel because they look parallel in a sketch, and setting consecutive interior angles equal when they should add to 180°.

Euclid’s first three postulates describe two tools. A straightedge draws a line through two points; it has no marks, so it cannot measure. A compass draws a circle or arc with a chosen center and radius. A construction is a drawing made with only those two tools. Start with copying a segment: draw a ray, set the compass to the length of the original segment, and swing an arc from the ray’s endpoint. Where the arc crosses the ray is the copy’s other endpoint. No ruler needed.

Next, the perpendicular bisector of segment AB. Open the compass wider than half of AB. Draw an arc from A, then an arc of the same radius from B. The arcs cross at two points, one above the segment and one below. Draw the line through those two points. Why does it work? Each crossing point is the same distance from A as from B, because both arcs used the same radius. A point equidistant from A and B lies on the perpendicular bisector, so the line through two such points is the bisector. It cuts AB at its midpoint and meets it at 90°.

To draw an angle bisector, swing an arc from the vertex V so it crosses both sides at points P and Q. Then draw equal arcs from P and Q that meet at a point R inside the angle. Ray VR splits the angle in half. Triangles VPR and VQR have three pairs of equal sides, which, as the next chapter proves, makes them congruent and their angles at V equal. To draw a line through a point parallel to a given line, copy an angle so that corresponding angles are congruent. Not everything is possible: mathematicians proved in the 1800s that no compass-and-straightedge construction can cut a general angle into three equal parts. The most common construction mistake is changing the compass width between two arcs that must match.

Slide a phone across a table without turning it. Every corner moves the same distance in the same direction. That motion is a translation. On a coordinate grid, a translation is written as a rule. The rule (x, y) → (x + 3, y − 2) means: add 3 to every x-coordinate and subtract 2 from every y-coordinate. Point (1, 5) moves to (4, 3). Point (−2, 0) moves to (1, −2). The original point is the preimage, and the moved point is the image. We mark an image with a prime: A moves to A′, read A prime.

Try a whole triangle. A(1, 2), B(4, 2) and C(1, 6), with the rule (x, y) → (x + 3, y − 2). Then A′ = (4, 0), B′ = (7, 0) and C′ = (4, 4). Now compare lengths. AB runs from x = 1 to x = 4 along the line y = 2, so AB = 3. A′B′ runs from x = 4 to x = 7, so A′B′ = 3. Every side keeps its length and every angle keeps its measure. A motion that preserves all distances and angles is a rigid motion. Translations, reflections and rotations are the three basic rigid motions, and this chapter builds the idea of congruence on them.

You can also read the rule off a pair of points. If a translation takes (2, 7) to (5, 3), it added 3 to x and subtracted 4 from y, so the rule is (x, y) → (x + 3, y − 4). The pair of numbers, 3 right and 4 down, is called the translation vector. Two common mistakes: subtracting when the rule says to add, and applying the change to only one coordinate. Every point gets both changes, every time.

Hold a letter up to a mirror and its image flips. A reflection flips every point across a line of reflection so that the line is the perpendicular bisector of the segment joining each point to its image. The point and its image are the same distance from the line, on opposite sides. A point on the line does not move at all. On the coordinate grid, three reflections have simple rules. Across the x-axis: (x, y) → (x, −y). Across the y-axis: (x, y) → (−x, y). Across the line y = x: (x, y) → (y, x).

Try the point (3, −5). Across the y-axis it becomes (−3, −5): the x-coordinate changes sign, and the point moves from 3 units right of the axis to 3 units left. Across the x-axis it becomes (3, 5). Across y = x, the coordinates swap, giving (−5, 3). Now reflect a triangle with vertices P(1, 1), Q(4, 1) and R(1, 3) across the y-axis: P′(−1, 1), Q′(−4, 1), R′(−1, 3). Side PQ has length 3, and so does P′Q′. A reflection is a rigid motion.

One thing does change. Read the original triangle’s vertices going around: P to Q to R runs counterclockwise. Read the image: P′ to Q′ to R′ runs clockwise. A reflection reverses orientation, the direction you travel around the figure. Translations and rotations keep orientation. That is why a mirror flips the writing on a shirt while a slide or a turn does not. Common mistake: changing the sign of the wrong coordinate. Across the x-axis, the point stays on the same vertical line, so x stays and y flips. Say it as: reflect across the x-axis, flip the y.

A clock hand turns around the center of the clock. A rotation turns every point of a figure around a fixed point, the center of rotation, through the same angle of rotation. Counterclockwise is the standard direction. On the coordinate grid, rotations about the origin have rules. A 90° counterclockwise rotation sends (x, y) to (−y, x). A 180° rotation sends (x, y) to (−x, −y). A 270° counterclockwise rotation, which is the same as a 90° clockwise rotation, sends (x, y) to (y, −x).

Follow the point (3, 2). After 90° counterclockwise it is at (−2, 3). Check that with a sketch: (3, 2) sits to the right and a bit up, and (−2, 3) sits to the left and up, a quarter turn around. After 180° the point is at (−3, −2), straight through the origin to the opposite side. After 270° it is at (2, −3). One more quarter turn, 360° in all, returns it to (3, 2). Each image is the same distance from the origin: √(9 + 4) = √13 every time, because a rotation is a rigid motion.

Rotate a triangle with vertices A(1, 1), B(4, 1) and C(4, 3) by 90° counterclockwise about the origin. Using (x, y) → (−y, x): A′(−1, 1), B′(−1, 4) and C′(−3, 4). Side AB was horizontal with length 3; side A′B′ is vertical with length 3. Rotations keep lengths, angles and orientation. The common mistakes are using the 90° rule when the problem says 180°, and forgetting to change the sign in the 90° rule. Write the rule down before you plug in any numbers.

Fold a paper heart down the middle and the two halves match. The fold line is a line of symmetry: a reflection across it maps the figure onto itself. A figure can have several. A rectangle that is not a square has two, one vertical and one horizontal. A square has four: the two through the midpoints of the sides and the two diagonals. A regular hexagon has six. A scalene triangle has none. The diagonals of a non-square rectangle are not lines of symmetry, which surprises people; fold along one and the corners stick out.

Rotational symmetry means a turn of less than 360° maps the figure onto itself. The smallest such turn is the angle of rotational symmetry. For a regular polygon with n sides, that angle is 360° ÷ n. A square: 360 ÷ 4 = 90°. A regular pentagon: 360 ÷ 5 = 72°. A regular hexagon: 360 ÷ 6 = 60°. The order of the symmetry is the number of positions that match in one full turn, which is n for a regular polygon. A non-square rectangle has rotational symmetry of 180° only, so its order is 2.

Letters make good practice. H has two lines of symmetry and 180° rotational symmetry. A has one vertical line of symmetry and no rotational symmetry. N has no line of symmetry, but turn it 180° and it looks the same, so it has rotational symmetry of order 2. Z is the same story. The common mistake is assuming any figure that looks balanced has both kinds of symmetry. Test each kind separately: fold for line symmetry, turn for rotational symmetry.

Two figures are congruent if a sequence of rigid motions maps one onto the other. The sequence can mix translations, reflections and rotations. That is the definition. It says something stronger than same size and same shape. It says you can physically move one figure until it sits exactly on top of the other. Rigid motions preserve lengths and angles, so congruent figures have equal corresponding parts. Every side matches a side of the same length. Every angle matches an angle of the same measure.

The order of the letters tells you which parts correspond. Triangle ABC ≅ triangle DEF means A matches D, B matches E and C matches F. So AB ≅ DE, BC ≅ EF and AC ≅ DF, and angle A ≅ angle D, angle B ≅ angle E, angle C ≅ angle F. Suppose triangle ABC ≅ triangle XYZ, with AB = 5, BC = 7 and angle A = 40°. Then XY = 5, YZ = 7 and angle X = 40°. You know six facts about triangle XYZ without ever seeing it. A congruence statement written in the wrong order gives wrong matches, so write it carefully.

Here is how to show congruence with motions. Triangle ABC has vertices A(0, 0), B(3, 0) and C(0, 4). Triangle DEF has vertices D(5, 2), E(8, 2) and F(5, 6). The translation (x, y) → (x + 5, y + 2) sends A to (5, 2) = D, B to (8, 2) = E, and C to (5, 6) = F. One rigid motion maps every vertex onto its partner, so the triangles are congruent. Sometimes you need two motions, such as a reflection followed by a translation. The common mistake is assuming two figures are congruent because they look alike. The test is a motion, or a criterion proved from motions, not a glance.

You do not need all six matching parts to know two triangles are congruent. The bridge truss showed the first shortcut. SSS (side-side-side): if three sides of one triangle are congruent to three sides of another, the triangles are congruent. That is exactly why the triangle frame would not fold. With sides 5, 7 and 9, there is only one triangle you can build, so any two triangles with those sides must match. The proof idea: translate one triangle so a side lands on its partner, then a reflection if needed, and the third vertex is forced because it must be the right distance from both ends.

The second shortcut is SAS (side-angle-side): two sides and the included angle, the angle between those two sides, congruent to the matching parts of another triangle. Picture two sticks of lengths 6 and 8 hinged at one end. Open the hinge to exactly 40°, and the distance between the far ends is set. Any other pair of 6 and 8 sticks opened to 40° will produce the same triangle. In triangle ABC, the angle included between sides AB and BC is angle B, because B is the letter the two sides share.

A typical proof. Given: AB ≅ CB and angle ABD ≅ angle CBD. Prove: triangle ABD ≅ triangle CBD. Statement 1: AB ≅ CB, given. Statement 2: angle ABD ≅ angle CBD, given. Statement 3: BD ≅ BD, reflexive property. Statement 4: triangle ABD ≅ triangle CBD by SAS, since the two given sides and the shared side surround the given angle. Common mistake: using an angle that is not between the two sides. Two sides and an angle somewhere else is a different situation, and a later lesson shows why it can fail.

A surveyor wants the distance across a river without crossing it. She marks two points A and B on her bank and measures AB. Then she measures the angle at A and the angle at B toward a tree T on the far bank. That is ASA (angle-side-angle): two angles and the included side, the side between them. Anyone who draws a segment of length AB and opens those two angles at its ends gets the same triangle. The two rays from A and B can meet in only one place. So the surveyor can draw a scale copy at her desk and read off the distance to the tree.

AAS (angle-angle-side) is two angles and a side that is not between them. It works because of the angle sum. If two angles of a triangle are 30° and 80°, the third must be 180° − 30° − 80° = 70°. So knowing two angles means knowing all three, and a side that was not included between the two given angles is included between one of them and the third. AAS turns into ASA every time. Suppose two triangles each have angles of 30° and 80° and a side of 12 opposite the 30° angle. That side lies between the 80° and 70° angles, so ASA applies with those two angles.

Notice what is missing from the list: AAA, three angles. Three angles do not force a size. A triangle with angles 60°, 60°, 60° and side 2 and another with side 10 have the same angles and are not congruent. They are the same shape at different sizes, which the next unit calls similar. Common mistake: choosing between ASA and AAS by guessing. Ask one question: is the given side between the two given angles? If yes, ASA. If no, AAS.

Take two sticks, one 8 cm and one 5 cm, hinged at a point B. Fix the 8 cm stick along a line and open a 30° angle at B. The 5 cm stick swings from the hinge, and its free end has to land somewhere on a ray that leaves the far end of the 8 cm stick at 30°. Set a compass to 5 cm and swing it from the far end of that ray: it crosses the ray in two places. That gives two different triangles with sides 8 and 5 and a 30° angle opposite the 5. Two sides and a non-included angle, called SSA, did not force one shape. That is why SSA is not on the list of criteria.

The numbers explain it. The shortest possible distance from the swinging stick’s pivot to the ray is 8 × sin 30° = 8 × 0.5 = 4 cm. Because 5 is bigger than 4 but smaller than 8, the compass arc reaches the ray twice. Make the swinging stick 4 cm and it touches once, making a right triangle. Make it 3 cm and it never reaches the ray, so no triangle exists. Make it 9 cm, longer than the fixed side, and it crosses the ray once again. So SSA sometimes gives two triangles, sometimes one, sometimes none. A criterion has to work every time.

One special case does work. In a right triangle, the hypotenuse is the side opposite the right angle, and the other two sides are legs. The HL criterion (hypotenuse-leg) says two right triangles with congruent hypotenuses and one pair of congruent legs are congruent. It looks like SSA, but the right angle rescues it: with hypotenuse 13 and leg 5, the other leg must satisfy 5^2 + b^2 = 13^2, so b^2 = 169 − 25 = 144 and b = 12. Every such triangle has sides 5, 12 and 13, so SSS finishes the job. Common mistake: writing SAS when the angle is not between the sides. Check where the angle sits before you name the criterion.

The congruence criteria are tools for something bigger. Prove two triangles congruent, and every pair of corresponding sides and angles comes free. That step has a name: CPCTC, corresponding parts of congruent triangles are congruent. It is the definition of congruence read backward. A proof that uses it has two halves: first prove the triangles congruent with SSS, SAS, ASA, AAS or HL, then use CPCTC to reach the side or angle you actually wanted.

Example. Given: AB ≅ CB, and BD bisects angle ABC. Prove: AD ≅ CD. Statement 1: AB ≅ CB, given. Statement 2: BD bisects angle ABC, given. Statement 3: angle ABD ≅ angle CBD, definition of angle bisector. Statement 4: BD ≅ BD, reflexive property. Statement 5: triangle ABD ≅ triangle CBD, SAS, because the angle at B is included between AB and BD in one triangle and between CB and BD in the other. Statement 6: AD ≅ CD, CPCTC. Notice that the goal, AD ≅ CD, was never assumed. It fell out of the congruent triangles.

Now attach numbers. In that figure, AD = 3x + 1 and CD = 5x − 7. Because AD ≅ CD, 3x + 1 = 5x − 7, so 8 = 2x and x = 4. Then AD = 13 and CD = 13. Surveyors use the same idea to measure across a pond. They pick a point P, walk equal distances past it on two lines to make two triangles with vertical angles at P, and the far side of the small triangle they can measure equals the width of the pond by SAS and CPCTC. Common mistake: using CPCTC before the triangles are proved congruent. It is always the step after.

A parallelogram is a quadrilateral with both pairs of opposite sides parallel. That is the whole definition. Everything else is a theorem proved with congruent triangles. Draw parallelogram ABCD, with vertices in order, and add the diagonal AC. Because AB is parallel to DC, the alternate interior angles BAC and DCA are congruent. Because AD is parallel to BC, the alternate interior angles DAC and BCA are congruent. AC is shared, so triangle ABC ≅ triangle CDA by ASA. CPCTC then gives AB ≅ CD and BC ≅ AD: opposite sides of a parallelogram are congruent.

The same triangles give angle B ≅ angle D, and drawing the other diagonal gives angle A ≅ angle C: opposite angles are congruent. Consecutive angles, like A and B, are same-side interior angles between parallel lines, so they are supplementary. If angle A measures 70°, then B = 110°, C = 70° and D = 110°. Try sides: AB = 2x + 3 and CD = 4x − 9. Opposite sides are congruent, so 2x + 3 = 4x − 9, giving x = 6 and AB = 15. The diagonals also bisect each other: if the diagonals cross at E, then AE = EC and BE = ED. If AE = 7, the full diagonal AC = 14.

The converses let you prove a quadrilateral is a parallelogram. Any one of these is enough: both pairs of opposite sides congruent, both pairs of opposite angles congruent, diagonals that bisect each other, or one pair of opposite sides that are both parallel and congruent. Common mistake: setting consecutive angles equal to each other. Opposite angles are equal; consecutive angles add to 180°. If a problem gives angle A = 3x and angle B = 2x + 30, write 3x + 2x + 30 = 180, not 3x = 2x + 30.

Three special parallelograms show up everywhere. A rectangle is a parallelogram with four right angles. A rhombus is a parallelogram with four congruent sides. A square is both: four right angles and four congruent sides. Each has every parallelogram property from the last lesson, plus something extra about its diagonals. In a rectangle, the diagonals are congruent. In a rhombus, the diagonals are perpendicular and each one bisects a pair of opposite angles. In a square, all of that is true at once.

The rhombus rule turns into a right-triangle problem. The diagonals of a rhombus measure 12 and 16. They bisect each other, so the four triangles they form have legs of 6 and 8. Each triangle is a right triangle because the diagonals are perpendicular. The side of the rhombus is the hypotenuse: 6^2 + 8^2 = 36 + 64 = 100, so the side is √100 = 10. All four sides are 10, as a rhombus requires. For a rectangle, use the congruent diagonals: if AC = 3x + 2 and BD = 5x − 8, then 3x + 2 = 5x − 8, so x = 5 and each diagonal is 17.

A quadrilateral with exactly one pair of parallel sides is a trapezoid, and if its non-parallel legs are congruent it is an isosceles trapezoid, whose base angles are congruent and whose diagonals are congruent. Keep the family tree straight: every square is a rectangle and a rhombus, every rectangle and rhombus is a parallelogram, but a rectangle need not be a rhombus. Common mistake: assuming a rectangle’s diagonals are perpendicular. Draw a long, thin rectangle and its diagonals cross at a sharp angle, not 90°. Only the rhombus and square get perpendicular diagonals.

Tear the three corners off a paper triangle and lay them side by side with their points touching. They form a straight line. That experiment is inductive reasoning, and it suggests the triangle angle sum theorem: the three interior angles of a triangle add to 180°. Here is the deductive proof. Take triangle ABC and draw a line through vertex A parallel to side BC. That line makes three angles at A that together form a straight angle, so they add to 180°. The middle one is angle A itself. The one on the left is congruent to angle B, because they are alternate interior angles for the parallel lines with transversal AB. The one on the right is congruent to angle C for the same reason with transversal AC. So angle A + angle B + angle C = 180°.

Now use it. A triangle has angles measuring 2x, 3x and 4x. Their sum is 9x = 180, so x = 20, and the angles are 40°, 60° and 80°. Check: 40 + 60 + 80 = 180. Another: two angles are 35° and 82°. The third is 180 − 35 − 82 = 63°. A quick consequence: a triangle can have at most one right angle or one obtuse angle, because two of them would already use up 180° or more.

The theorem also classifies triangles. An acute triangle has three angles under 90°. A right triangle has one 90° angle, and its other two angles must add to 90°, so they are complementary. An obtuse triangle has one angle over 90°. Common mistakes: forgetting that the sum is 180° and not 360°, which is the sum for a quadrilateral, and dropping one of the angles when solving. Write all three angles, add them, and set the sum equal to 180 every time.

Extend one side of a triangle past a vertex. The angle between the extension and the next side is an exterior angle. It forms a linear pair with the interior angle at that vertex, so the two add to 180°. The two interior angles that are not at that vertex are the remote interior angles. The exterior angle theorem says the exterior angle equals the sum of the two remote interior angles. Proof: interior angle C plus the exterior angle at C is 180° (linear pair). Interior angles A + B + C are also 180° (angle sum). Both sums equal 180°, so subtract angle C from each: the exterior angle equals A + B.

Try it with numbers. A triangle has remote interior angles of 48° and 71°. The exterior angle at the third vertex is 48 + 71 = 119°. Check with the long route: the third interior angle is 180 − 48 − 71 = 61°, and the exterior angle is 180 − 61 = 119°. Both routes agree. Now with algebra: an exterior angle measures (4x + 10)°, and the remote interior angles are (2x + 5)° and (x + 35)°. Set 4x + 10 = 2x + 5 + x + 35, which is 4x + 10 = 3x + 40, so x = 30. The exterior angle is 130°, and the remote interior angles are 65° and 65°. Check: the interior angle at that vertex is 50°, and 50 + 130 = 180.

Why does this shortcut matter? It saves a step, and it gives an inequality for free: an exterior angle is larger than either remote interior angle alone, since it equals their sum and both are positive. That fact returns later in this chapter when we compare sides and angles. Common mistake: adding the adjacent interior angle instead of the two remote ones. The exterior angle and its neighbor add to 180°; the exterior angle and the two far angles are equal.

An isosceles triangle has at least two congruent sides, called legs. The third side is the base, and the two angles at the ends of the base are the base angles. The isosceles triangle theorem, sometimes called the base angles theorem, says the base angles are congruent. Proof: in triangle ABC with AB ≅ AC, draw the bisector of angle A and let it meet BC at D. Then AB ≅ AC (given), angle BAD ≅ angle CAD (definition of bisector) and AD ≅ AD (reflexive). Triangle ABD ≅ triangle ACD by SAS, so angle B ≅ angle C by CPCTC.

The converse also holds: if two angles of a triangle are congruent, the sides opposite them are congruent. Numbers make the theorem useful. If the vertex angle of an isosceles triangle is 40°, the two base angles share the remaining 140°, so each is 70°. If a base angle is 55°, the other base angle is 55° and the vertex angle is 180 − 110 = 70°. With algebra: the base angles are (3x + 5)° and (5x − 21)°. They are congruent, so 3x + 5 = 5x − 21, giving 26 = 2x and x = 13. Each base angle is 44°, and the vertex angle is 180 − 88 = 92°.

An equilateral triangle has three congruent sides. Apply the theorem twice and all three angles are congruent, so each is 180 ÷ 3 = 60°. The converse works too: a triangle with three 60° angles is equilateral. Common mistake: mixing up which angle is the vertex angle. The vertex angle is between the two legs; the base angles are the equal pair. If a problem says one angle of an isosceles triangle is 100°, that angle must be the vertex angle, because two 100° base angles would exceed 180°.

The fire station surveyor used one fact over and over. The perpendicular bisector theorem says: if a point lies on the perpendicular bisector of a segment, it is equidistant from the segment’s endpoints. Proof sketch: let the bisector cross AB at midpoint M, and let P be any point on the bisector. Then AM ≅ MB, the angles at M are right angles, and PM ≅ PM, so triangle PMA ≅ triangle PMB by SAS and PA ≅ PB by CPCTC. The converse is also true: any point equidistant from A and B lies on the perpendicular bisector.

Use it with algebra. Point P lies on the perpendicular bisector of AB, with PA = 4x − 1 and PB = 2x + 9. Since PA = PB, 4x − 1 = 2x + 9, so 2x = 10 and x = 5. Then PA = 19 and PB = 19. Now the theorem about triangles: the three perpendicular bisectors of a triangle’s sides are concurrent, meaning they all pass through one point. That point is the circumcenter. It is equidistant from the three vertices, so a circle centered there passes through all three: the circumscribed circle. The chapter story’s fire station was the circumcenter.

Where the circumcenter sits depends on the triangle. For an acute triangle it is inside. For a right triangle it is exactly at the midpoint of the hypotenuse. For an obtuse triangle it is outside, which is what put the fire station in the cornfield. Try a right triangle with vertices at (0, 0), (8, 0) and (0, 6). The hypotenuse runs from (8, 0) to (0, 6), with midpoint (4, 3). Distance from (4, 3) to (0, 0) is √(16 + 9) = 5; to (8, 0) it is √(16 + 9) = 5; to (0, 6) it is √(16 + 9) = 5. All equal, so (4, 3) is the circumcenter. Common mistake: confusing equidistant from the vertices with equidistant from the sides. That second one is the next lesson.

The county board’s second idea was a station equally far from the three highways. Distance from a point to a line always means the perpendicular distance, the shortest path. The angle bisector theorem says: if a point lies on the bisector of an angle, it is equidistant from the two sides of the angle. Proof sketch: from point P on the bisector, drop perpendiculars to the two sides, landing at X and Y. Angle XVP ≅ angle YVP (bisector), the angles at X and Y are right angles, and VP ≅ VP, so triangle VXP ≅ triangle VYP by AAS and PX ≅ PY by CPCTC. The converse holds too.

With algebra: P is on the bisector of angle V, and its distances to the two sides are 2x + 3 and 4x − 7. Set 2x + 3 = 4x − 7, so 10 = 2x and x = 5. Each distance is 13. In a triangle, the three angle bisectors are concurrent at the incenter. The incenter is equidistant from the three sides, so a circle centered there touches all three sides from the inside: the inscribed circle. That circle is the largest one that fits in the triangle, and its radius is the incenter’s distance to any side.

Unlike the circumcenter, the incenter is always inside the triangle, because every angle bisector runs through the triangle’s interior. That is why it answers the board’s second question so well: a station at the incenter is equally close to all three highways and never ends up in a cornfield. Common mistake: measuring distance from a point to a line along a slanted path. Only the perpendicular counts. If a problem gives a slanted segment, it is not the distance, and the theorem does not apply to it.

A median of a triangle is a segment from a vertex to the midpoint of the opposite side. Every triangle has three, and they are concurrent at a point called the centroid. The centroid has a precise location: it lies two thirds of the way along each median, measured from the vertex. If AD is a median with centroid G, then AG = 2/3 AD and GD = 1/3 AD, so AG is twice GD. This is the balance point the county board wondered about. Cut a triangle from stiff cardboard and it will balance on a pencil tip placed at the centroid.

Numbers first. Median AD is 18 units long. Then AG = 2/3 × 18 = 12 and GD = 18 − 12 = 6. Check: 12 is twice 6. Going the other way: if GD = 5, then AG = 10 and the whole median AD = 15. With algebra, suppose AG = 4x and GD = x + 3. Since AG = 2 GD, 4x = 2(x + 3) = 2x + 6, so 2x = 6 and x = 3. Then AG = 12, GD = 6 and AD = 18.

Why two thirds? The two medians from B and C cut each other in the same ratio, and a proof using midsegments, which appear in the next lesson, shows that the crossing point sits at the two-thirds mark on each. The centroid is always inside the triangle, since each median runs through the interior. Common mistake: using 1/2 instead of 2/3. A median is not bisected at the centroid. The vertex piece is the long piece, always twice the other.

An altitude of a triangle is the perpendicular segment from a vertex to the line containing the opposite side. It is the height you use in the area formula. The three altitudes, or the lines that contain them, meet at the orthocenter. In an acute triangle the orthocenter is inside. In a right triangle two altitudes are the legs themselves, so the orthocenter is at the right-angle vertex. In an obtuse triangle two altitudes fall outside the triangle, and the orthocenter is outside. Do not confuse an altitude with a median: the altitude is perpendicular to the side, while the median goes to its midpoint. They coincide only in special triangles such as the equilateral triangle.

A midsegment joins the midpoints of two sides of a triangle. The midsegment theorem says a midsegment is parallel to the third side and half as long. Suppose D and E are the midpoints of AB and AC. Then DE is parallel to BC and DE = 1/2 BC. If BC = 14, then DE = 7. If DE = 7, then BC = 14. With algebra: DE = 2x + 1 and BC = 6x − 8. Because BC = 2 DE, write 6x − 8 = 2(2x + 1) = 4x + 2, so 2x = 10 and x = 5. Then DE = 11 and BC = 22. Check: 22 is twice 11.

The three midsegments cut a triangle into four congruent triangles, and the middle one, the midsegment triangle, has half the perimeter of the original. If a triangle has sides 10, 12 and 16, the midsegments are 5, 6 and 8, with perimeter 19, half of 38. Common mistake: doubling the wrong segment. The midsegment is the short one. When you see DE = 3x and BC = 4x + 10, the equation is 4x + 10 = 2(3x), which gives 4x + 10 = 6x and x = 5, so BC = 30 and DE = 15.

Walk from the Bean in Millennium Park to the Art Institute, then on to the Harold Washington Library. Or walk straight from the Bean to the library. The two-leg route is always longer than the straight route, because a straight segment is the shortest path between two points. That everyday fact is the triangle inequality theorem: the sum of the lengths of any two sides of a triangle is greater than the length of the third side. If it were not, the two short sides could not reach each other across the long side, and the triangle would not close.

Test some side lengths. Sides 5, 7 and 11: 5 + 7 = 12, and 12 > 11, so a triangle exists. You only need to check the two smallest sides against the largest, because the other two sums are automatically bigger. Sides 4, 5 and 10: 4 + 5 = 9, and 9 is not greater than 10, so no triangle. Sides 4, 4 and 8: 4 + 4 = 8, which equals 8 but is not greater, so no triangle. The two short sides would lie flat along the long one.

The theorem also gives a range for a missing side. Two sides are 6 and 10. The third side x must satisfy x < 6 + 10 = 16 and x > 10 − 6 = 4. So 4 < x < 16. In words: the third side is less than the sum and greater than the difference of the other two. With sides 3 and 5, the third side is between 2 and 8, so the whole-number possibilities are 3, 4, 5, 6 and 7: five of them. Common mistake: writing the range with ≤ instead of <. The endpoints give a flat, closed-up figure, not a triangle.

Open a door a little, and the gap between its edge and the frame is small. Open it wide, and the gap grows. The door, the frame and the gap form a triangle, and the wider angle at the hinge stretches the side across from it. That is the idea behind two inequality theorems. First: if one side of a triangle is longer than another, the angle opposite the longer side is larger than the angle opposite the shorter side. Second, its converse: if one angle is larger than another, the side opposite the larger angle is longer. In short, biggest angle faces longest side, and smallest angle faces shortest side.

Try it. A triangle has angle A = 50°, angle B = 70° and angle C = 60°. The largest angle is B, so the longest side is the one opposite B, which is AC. The smallest angle is A, so the shortest side is BC. Going the other way: sides AB = 9, BC = 5 and AC = 7. The shortest side BC is opposite angle A, so angle A is the smallest angle. The longest side AB is opposite angle C, so angle C is the largest. You can order all three angles without measuring one of them.

The door idea has a name: the hinge theorem. If two triangles have two pairs of congruent sides, and the included angle of one is larger than the included angle of the other, then the third side of the first triangle is longer. Two triangles each have sides 6 and 8. One has a 50° angle between them, the other 65°. The 65° triangle has the longer third side. Its converse also holds: a longer third side means a larger included angle. Common mistake: matching the largest angle with the side next to it instead of the side across from it. Opposite means across, never adjacent.

A coordinate proof places a figure on the coordinate plane and proves a statement with formulas. Three tools do the work. The distance formula: the distance between (x1, y1) and (x2, y2) is √((x2 − x1)^2 + (y2 − y1)^2). The midpoint formula: the midpoint is ((x1 + x2)/2, (y1 + y2)/2). And slope: (y2 − y1)/(x2 − x1). Equal slopes mean parallel segments. Slopes whose product is −1 mean perpendicular segments. Place the figure smartly: put one vertex at the origin and one side along the x-axis so many coordinates are 0.

Prove that the triangle with vertices A(0, 0), B(6, 0) and C(3, 4) is isosceles. AC = √((3 − 0)^2 + (4 − 0)^2) = √(9 + 16) = √25 = 5. BC = √((3 − 6)^2 + (4 − 0)^2) = √(9 + 16) = 5. Two sides are congruent, so the triangle is isosceles. Now check the midsegment theorem on the same triangle. The midpoint of AC is (1.5, 2) and the midpoint of BC is (4.5, 2). The midsegment joining them has length 4.5 − 1.5 = 3, which is half of AB = 6, and its slope is 0, the same as AB. Parallel and half as long, exactly as the theorem promised.

A right angle can be checked two ways. The triangle with vertices (1, 1), (4, 1) and (4, 5) has a horizontal side of length 3 and a vertical side of length 4, and the slanted side is √(9 + 16) = 5. Since 3^2 + 4^2 = 9 + 16 = 25 = 5^2, the triangle is a right triangle. Or notice that the horizontal side has slope 0 and the vertical side has undefined slope, so they are perpendicular. Common mistakes: forgetting to square the differences in the distance formula, and forgetting the square root at the end. Write the formula every time before you plug in.

Coordinates make two centers easy to compute. The centroid of a triangle is the average of its three vertices: add the three x-coordinates and divide by 3, then do the same for y. For the triangle with vertices (0, 0), (6, 0) and (3, 9), the centroid is ((0 + 6 + 3)/3, (0 + 0 + 9)/3) = (3, 3). Check it against the two-thirds rule. The median from (3, 9) goes to the midpoint of the base, (3, 0). It is vertical with length 9. Two thirds of the way down from (3, 9) is 9 − 6 = 3, so the point (3, 3) is on that median at the right spot.

For a right triangle, the circumcenter is the midpoint of the hypotenuse. Take the triangle with vertices (0, 0), (10, 0) and (0, 4). The right angle is at the origin. So the hypotenuse joins (10, 0) and (0, 4), and its midpoint is (5, 2). Verify it. The distance from (5, 2) to (0, 0) is √(25 + 4) = √29. To (10, 0) it is √(25 + 4) = √29. To (0, 4) it is √(25 + 4) = √29. All three match. So (5, 2) is equidistant from the vertices, exactly what a circumcenter must be. The radius of the circumscribed circle is √29.

For a triangle that is not a right triangle, write equations for two perpendicular bisectors. The circumcenter is where they cross. The chapter story did this. For towns at (0, 0), (8, 0) and (0, 6), the bisectors x = 4 and y = 3 meet at (4, 3). Two common mistakes: averaging only two vertices for the centroid, and using the midpoint of a leg instead of the hypotenuse. The centroid needs all three vertices, and the circumcenter shortcut needs the side opposite the right angle.