Geometric Proofs

A **geometric proof** shows that a fact about a figure must be true. It is a chain of statements, and every statement has a reason: the given information, a definition, a postulate or a theorem that is already proven. A drawing helps you find the idea, but a measurement with a protractor is never a reason. The toolbox for this lesson: **1. Parallel lines and a transversal.** A **transversal** is a line that crosses two other lines. If the two lines are parallel: - **Alternate interior angles** (between the parallel lines, on opposite sides of the transversal) are equal. - **Corresponding angles** (in the same position at both crossings) are equal. - Same-side interior angles add up to $180^\circ$. **2. Triangle angle sum.** In every triangle $\angle A + \angle B + \angle C = 180^\circ$. **3. Congruent triangles.** Two triangles are congruent ($\cong$) when one of these sets of matching parts is equal: - **SSS**: three sides - **SAS**: two sides and the angle between them - **ASA**: two angles and the side between them - **AAS**: two angles and a side that is not between them **4. CPCTC.** Corresponding Parts of Congruent Triangles are Congruent: once you know $\triangle ABC \cong \triangle DEF$, every matching side and angle is equal. **5. Isosceles triangles.** If two sides of a triangle are equal, the angles opposite them (the **base angles**) are equal. **6. Midpoints and bisectors.** A **midpoint** splits a segment into two equal parts, an **angle bisector** splits an angle into two equal angles. A complete proof in two columns. Given: segments $AD$ and $BC$ cross at $M$, and $M$ is the midpoint of both. Prove: $AB = DC$. | Statement | Reason | |---|---| | $AM = DM$ | Definition of midpoint | | $BM = CM$ | Definition of midpoint | | $\angle AMB = \angle DMC$ | Vertical angles are equal | | $\triangle AMB \cong \triangle DMC$ | SAS | | $AB = DC$ | CPCTC | The usual plan: find two triangles that contain the part you want, prove them congruent, then finish with CPCTC.

Angles at Two Parallel Lines

Lines $AB$ and $CD$ are parallel. Points $A$ and $C$ are on the left, $B$ and $D$ on the right. A transversal crosses $AB$ at $E$ and $CD$ at $F$, and $\angle AEF = 65^\circ$. Find $\angle EFD$ and $\angle CFE$ and justify each step.

Locate the angle pair: $\angle AEF$ and $\angle EFD$ both lie between the parallel lines, on opposite sides of the transversal = They are alternate interior angles

Apply the parallel line theorem: Alternate interior angles at parallel lines are equal, so $\angle EFD = \angle AEF$ = $\angle EFD = 65^\circ$

Use the straight angle at $F$: $\angle CFE$ and $\angle EFD$ together form the straight angle $CFD$: $\angle CFE = 180^\circ - 65^\circ$ = $\angle CFE = 115^\circ$ (linear pair)

Answer: $\angle EFD = 65^\circ$ (alternate interior angles) and $\angle CFE = 115^\circ$ (linear pair). Check: $\angle AEF + \angle CFE = 180^\circ$, because same-side interior angles add up to $180^\circ$.

Proving the Triangle Angle Sum

Prove that the angles of any triangle $ABC$ add up to $180^\circ$. Draw the line $DE$ through $A$ parallel to $BC$, with $D$ on the side of $B$ and $E$ on the side of $C$.

Statement 1: the straight angle at $A$: $D$, $A$ and $E$ lie on one line, so $\angle DAB + \angle BAC + \angle CAE = 180^\circ$ = Reason: angles on a straight line

Statement 2: transversal $AB$: $DE \parallel BC$ and $AB$ crosses both lines, so $\angle DAB = \angle ABC$ = Reason: alternate interior angles

Statement 3: transversal $AC$: $AC$ also crosses both parallel lines, so $\angle CAE = \angle ACB$ = Reason: alternate interior angles

Statement 4: substitute: Replace $\angle DAB$ and $\angle CAE$ in statement 1 using statements 2 and 3 = $\angle ABC + \angle BAC + \angle ACB = 180^\circ$

Answer: The parallel line moves the three angles of the triangle side by side into the straight angle at $A$, so they add up to $180^\circ$ in every triangle. Q.E.D.

SSS and CPCTC in a Kite

In quadrilateral $ABCD$, $AB = AD$ and $CB = CD$. Prove that the diagonal $AC$ bisects $\angle BAD$. Then find $\angle BAC$ if $\angle BAD = 70^\circ$.

List the pairs of sides: $AB = AD$ (given), $CB = CD$ (given), $AC = AC$ (reflexive property) = Three pairs of equal sides

Prove the triangles congruent: Match the vertices $A \to A$, $B \to D$, $C \to C$ = $\triangle ABC \cong \triangle ADC$ (SSS)

Use CPCTC: $\angle BAC$ and $\angle DAC$ are corresponding angles of the congruent triangles = $\angle BAC = \angle DAC$ (CPCTC)

Conclude and compute: Two equal halves mean that $AC$ bisects $\angle BAD$: $\angle BAC = 70^\circ \div 2$ = $\angle BAC = 35^\circ$

Answer: $AC$ bisects $\angle BAD$ (SSS, then CPCTC), and $\angle BAC = \angle DAC = 35^\circ$.

Base Angles of an Isosceles Triangle

Given: $AB = AC$ in $\triangle ABC$. Prove: $\angle B = \angle C$. Then find the base angles if $\angle A = 40^\circ$.

Add a helper line: Draw the bisector of $\angle BAC$, it meets $BC$ at $D$. Then $\angle BAD = \angle CAD$ = Reason: definition of angle bisector

Collect the parts for SAS: $AB = AC$ (given), $\angle BAD = \angle CAD$ (step 1), $AD = AD$ (reflexive property) = Two sides and the angle between them

Prove the triangles congruent: The angle at $A$ lies between the sides $AB$, $AD$ and between $AC$, $AD$ = $\triangle ABD \cong \triangle ACD$ (SAS)

Use CPCTC: $\angle B$ and $\angle C$ are corresponding angles = $\angle B = \angle C$ (CPCTC)

Compute with $\angle A = 40^\circ$: $\angle B + \angle C = 180^\circ - 40^\circ = 140^\circ$, and the two angles are equal = $\angle B = \angle C = 70^\circ$

Answer: The base angles of an isosceles triangle are equal. With $\angle A = 40^\circ$ each base angle is $70^\circ$. CPCTC also gives $BD = CD$, so $D$ is the midpoint of $BC$.

Parallel Lines, ASA and a Midpoint

Segments $AE$ and $BD$ cross at $C$. $AB \parallel DE$ and $C$ is the midpoint of $AE$. Prove that $C$ is also the midpoint of $BD$. Then find $\angle CDE$ if $\angle BAC = 50^\circ$ and $\angle ACB = 70^\circ$.

Angles at $A$ and $E$: $AB \parallel DE$, and $AE$ is a transversal = $\angle BAC = \angle DEC$ (alternate interior angles)

Sides at $C$: $C$ is the midpoint of $AE$ = $AC = EC$ (definition of midpoint)

Angles at $C$: $\angle ACB$ and $\angle ECD$ are opposite each other = $\angle ACB = \angle ECD$ (vertical angles)

Congruence and CPCTC: The side $AC$ lies between the angles at $A$ and $C$, so $\triangle ABC \cong \triangle EDC$ (ASA) = $BC = DC$ (CPCTC), so $C$ is the midpoint of $BD$

Find the angle: $\angle CDE = \angle CBA = 180^\circ - 50^\circ - 70^\circ$ (CPCTC and angle sum) = $\angle CDE = 60^\circ$

Answer: $C$ is the midpoint of $BD$ (ASA, then CPCTC), and $\angle CDE = 60^\circ$.

Geometric proofs are where mathematical proof was born, and their ideas are still used every day: - **History of mathematics**: Euclid's Elements (around 300 BC) built all of geometry from a few postulates with exactly these proofs, and it was a school textbook for more than 2000 years - **Surveying**: Congruent triangles let surveyors measure distances they cannot walk, like the width of a river or a pond - **Architecture and engineering**: Roof trusses and bridges use isosceles and congruent triangles, because a triangle with fixed sides cannot change shape - **Road and city planning**: Parallel streets crossed by a diagonal avenue repeat the same angles at every corner - **Reasoning skills**: Proving that something must be true, step by justified step, is the same skill lawyers, programmers and scientists need Once you know these few theorems well, a large part of school geometry turns into short, convincing arguments.

Mistake: Using AAA (three equal angles) to prove that triangles are congruent

Why: Equal angles only give the same shape: a small and a large equilateral triangle have the same angles but are not congruent. Two sides and an angle that is not between them (SSA) also do not work in general, because two different triangles can fit.

Correct: Use only SSS, SAS, ASA or AAS, and in SAS check that the angle lies between the two sides.

Mistake: Writing CPCTC before the triangles are proven congruent

Why: CPCTC is only allowed after a line that says the triangles are congruent. Otherwise the proof assumes what it wants to show.

Correct: First a line like $\triangle ABC \cong \triangle DEF$ (SAS), then the matching sides or angles with the reason CPCTC.

Mistake: Using angle facts for parallel lines when the lines are not parallel, or saying same-side interior angles are equal

Why: Alternate interior and corresponding angles are equal only at parallel lines. Same-side interior angles are not equal, they add up to $180^\circ$.

Correct: Check that the lines are given as parallel ($AB \parallel CD$), then name the pair exactly: alternate interior (equal), corresponding (equal) or same-side interior (sum $180^\circ$).

Mistake: Writing the vertices of a congruence statement in the wrong order

Why: $\triangle ABC \cong \triangle DEF$ means $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$. A wrong order makes CPCTC pair the wrong sides and angles.

Correct: Match the vertices by the equal parts. In Example 5 the correct statement is $\triangle ABC \cong \triangle EDC$, not $\triangle ABC \cong \triangle DEC$.

Roof Trusses

Carpenters build roof trusses as isosceles triangles: two rafters of equal length meet at the ridge, so both roof slopes have the same angle.

A truss with a $100^\circ$ angle at the ridge has base angles of $(180^\circ - 100^\circ) \div 2 = 40^\circ$.

Measuring Across a Pond

Surveyors measure distances they cannot walk with congruent triangles, using the same SAS and CPCTC proof as in this lesson.

To find the width $AB$ of a pond, walk from $A$ to a point $C$ and the same distance further to $D$. Then walk from $B$ through $C$ the same distance to $E$. Now $\triangle ACB \cong \triangle DCE$ (SAS), so $DE = AB$ by CPCTC.

Street Maps

Parallel streets crossed by a diagonal avenue form the angle pattern of parallel lines and a transversal, which city planners use to design corners and crossings.

If a diagonal avenue makes a $58^\circ$ angle with one street, it makes the same $58^\circ$ corresponding angle with every parallel street.

Every statement in a geometric proof needs a reason: given, definition, postulate or theorem

At parallel lines cut by a transversal, alternate interior and corresponding angles are equal, and same-side interior angles add up to $180^\circ$

The triangle angle sum ($180^\circ$) is proved with a line through one vertex parallel to the opposite side

Triangles are congruent by SSS, SAS, ASA or AAS, never by AAA and in general not by SSA

CPCTC: after proving two triangles congruent, all their corresponding sides and angles are equal

In an isosceles triangle the base angles are equal, and midpoints and bisectors give equal segments and equal angles

Q: What does CPCTC stand for?

A: Corresponding Parts of Congruent Triangles are Congruent. It is the reason you write after proving two triangles congruent, when you conclude that a matching side or angle is equal.

Q: Why is there no SSA criterion?

A: With two sides and an angle that is not between them, the third side can sometimes swing into two different positions, which gives two different triangles. So SSA does not guarantee congruence in general.

Q: What is the difference between ASA and AAS?

A: In ASA the known side lies between the two angles, in AAS it does not. Both work: two angles fix the third angle (angle sum $180^\circ$), so AAS can always be turned into ASA.

Q: How do I find which triangles to use?

A: Look at the segment or angle you must prove equal and find two triangles that contain it. Mark everything you know: given parts, vertical angles, a shared side. If you reach SSS, SAS, ASA or AAS, finish with CPCTC.

Exit Presentation

Geometric Proofs

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Logic & Sets / Mathematical Proofs

Geometric Proofs

Prove geometric facts step by step with parallel line angles, the triangle angle sum, triangle congruence (SSS, SAS, ASA, AAS), CPCTC and isosceles triangles.

Advanced30 min

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