Lesson 3: Triangles and Congruence

Triangles are the atoms of geometry: every polygon can be split into them, and every rigid structure depends on them. This lesson answers a practical question — when do two triangles match exactly? — with a short list of congruence postulates. Mastering them (and knowing the two impostors that do *not* work) unlocks most of the proof writing in this course.

Angle facts inside a triangle

The Triangle Angle-Sum Theorem: the interior angles of a triangle sum to 180°. The Exterior Angle Theorem: an exterior angle of a triangle equals the sum of the two non-adjacent (remote) interior angles — and is therefore greater than either of them alone.

The congruence postulates

Two triangles are congruent when all corresponding sides and angles match. You never need to check all six parts; any of these shortcuts is enough:

• SSS: three sides of one triangle congruent to three sides of the other.
• SAS: two sides and the included angle (the angle between those two sides).
• ASA: two angles and the included side.
• AAS: two angles and a non-included side.
• HL (right triangles only): the hypotenuse and one leg.

Once congruence is established, CPCTC — Corresponding Parts of Congruent Triangles are Congruent — lets you conclude any remaining part matches. CPCTC is the usual *goal* of a proof: you prove the triangles congruent in order to harvest one extra fact.

Two combinations do not prove congruence: SSA (two sides and a non-included angle — the "swinging door" can make two different triangles) and AAA (proves similarity, Lesson 4, but sizes can differ).

Worked example 1: Angle algebra

A triangle has angles measuring (2x)°, (3x + 10)° and (x + 50)°. Sum to 180°: 2x + 3x + 10 + x + 50 = 180, so 6x + 60 = 180 and x = 20. The angles are 2(20) = 40°, 3(20) + 10 = 70°, and 20 + 50 = 70°. Check: 40 + 70 + 70 = 180°. ✓ It is an isosceles triangle with two 70° angles.

Worked example 2: Proving congruence with coordinates (SSS)

Plot △ABC with A(0, 0), B(4, 0), C(0, 3) and △DEF with D(1, 1), E(5, 1), F(1, 4). Compute the sides. For △ABC: AB = 4 (horizontal), AC = 3 (vertical), BC = √(4² + 3²) = √25 = 5. For △DEF: DE = 4, DF = 3, EF = √(4² + 3²) = 5. All three pairs match: AB ≅ DE, AC ≅ DF, BC ≅ EF. By SSS, △ABC ≅ △DEF. Note the vertex order matters: A↔D, B↔E, C↔F — the congruence statement itself records the correspondence.

Worked example 3: A proof using SAS and vertical angles

Two segments, AC and BD, cross at point E, and E is the midpoint of both segments. Prove that △AEB ≅ △DEC:

Statements                                Reasons
1. E is the midpoint of AC and of BD     1. Given
2. AE ≅ CE and BE ≅ DE                   2. Definition of midpoint
3. ∠AEB ≅ ∠DEC                           3. Vertical angles are congruent
4. △AEB ≅ △DEC                           4. SAS (steps 2, 3, 2)
5. ∠BAE ≅ ∠DCE                           5. CPCTC

Step 5 is the point of the exercise: those congruent alternate interior angles (for transversal AC cutting lines AB and DC) prove AB ∥ DC by the converse from Lesson 2. Proofs chain lessons together like this all the time.

Geometry at work

A triangle is the only polygon that is automatically rigid: fix its three side lengths and its shape cannot flex (that is SSS at work). Rectangles, by contrast, can lean into parallelograms without any side changing length. That is why bridges, cranes, roof trusses and bicycle frames are full of triangles, and why a wobbly square gate gets fixed by adding one diagonal board — splitting it into two triangles.

Common mistakes

• Using SSA. Two sides and a non-included angle do not guarantee congruence. If the angle is not between the two sides, stop and look for another route.
• Sloppy correspondence. In △ABC ≅ △DEF, side AB corresponds to DE. Writing the statement out of order makes CPCTC conclusions wrong.
• Claiming CPCTC before proving congruence. CPCTC is only valid *after* a postulate has established the triangles are congruent.
• Counting the diagram instead of the givens. Hash marks and right-angle marks are data; lengths that merely *look* equal are not.

Exam Experience

1. Two triangles have two sides and the angle *between* those sides congruent. Which postulate proves them congruent?

• A) SSS
• B) SAS
• C) AAS
• D) SSA

Answer: B. The angle is included, so SAS applies. (SSA is not a valid postulate at all.)

2. An exterior angle of a triangle measures 130°. One remote interior angle measures 55°. What is the other remote interior angle?

• A) 50°
• B) 65°
• C) 75°
• D) 125°

Answer: C. Exterior Angle Theorem: 130° − 55° = 75°.

3. In right triangles, the HL postulate requires which two parts to be congruent?

• A) Two legs
• B) The hypotenuse and one leg
• C) Two acute angles and a leg
• D) The hypotenuse and one acute angle

Answer: B. HL uses the hypotenuse and either leg, and works only for right triangles.

Constructed response. In quadrilateral ABCD, side AB is parallel to side CD, and AB ≅ CD. Segment AC is a diagonal shared by triangles △ABC and △CDA. Prove △ABC ≅ △CDA, and state what CPCTC then tells you about BC and DA.

Scoring guide (4 points):

• 1 point: States ∠BAC ≅ ∠DCA as alternate interior angles formed by the parallel sides AB and CD cut by transversal AC.
• 1 point: States AC ≅ CA by the reflexive property.
• 1 point: Notes that ∠BAC is the angle included between sides AB and AC in △ABC, and ∠DCA is the angle included between sides CD and CA in △CDA, so with AB ≅ CD (given) the SAS postulate applies.
• 1 point: Concludes △ABC ≅ △CDA by SAS and, by CPCTC, BC ≅ DA.

Going deeper: Base angles and the exterior angle with algebra

In an isosceles triangle, the angles at the base (the side that is *not* one of the congruent pair) are the base angles, and they are congruent; the angle between the congruent sides is the vertex angle. If the vertex angle of an isosceles triangle is 40°, the base angles share what is left equally: (180° − 40°)/2 = 70° each. Conversely, if one base angle is 65°, the other is 65° and the vertex angle is 180° − 130° = 50°. These two computations — vertex to bases, base to vertex — are the bread and butter of triangle angle chasing.

Worked example 4: Exterior angle algebra

An exterior angle of a triangle measures 120°. The two remote interior angles measure (x + 30)° and (2x)°. By the Exterior Angle Theorem, (x + 30) + 2x = 120, so 3x + 30 = 120 and x = 30. The remote interiors are 30 + 30 = 60° and 2(30) = 60°, and indeed 60 + 60 = 120. ✓ The third angle of the triangle is supplementary to the exterior angle: 180° − 120° = 60°, making this an equiangular (hence equilateral) triangle in disguise. Angle chasing often ends with a recognition like that — the numbers announce what the figure really is.

The Triangle Inequality: which triples of sides can exist?

Not every three lengths form a triangle. The Triangle Inequality Theorem says the sum of any two sides must exceed the third — otherwise the two short sides cannot reach each other. Test 4, 6, 11: 4 + 6 = 10, which is less than 11, so no triangle. Test 5, 7, 11: 5 + 7 = 12 > 11, and the other two checks (5 + 11 > 7, 7 + 11 > 5) are automatic when the two smallest sum past the largest — so a triangle exists (barely; it will be very flat). The theorem also gives ranges: if two sides are 6 and 10, the third side t satisfies 10 − 6 < t < 10 + 6, i.e. 4 < t < 16. The lower bound comes from the same inequality rearranged (t + 6 > 10). Carpenters meet this when a brace length must be chosen before the frame is assembled; exam writers meet it whenever they want a "which of these cannot be a triangle?" question. It is also the deep reason the reflection trick of Lesson 19 works: a straight path beats any bent path because of the triangle inequality applied point by point.

Final drill: One more CPCTC harvest

Prove that the base angles of an isosceles triangle are congruent using a midpoint — but watch how the plan is chosen. In △PQR, PQ ≅ PR, and S is the midpoint of QR. Plan: show △PQS ≅ △PRS. The parts: PQ ≅ PR (given); QS ≅ RS (S is the midpoint, by definition); PS ≅ PS (reflexive). That is SSS, so the triangles are congruent, and ∠Q ≅ ∠R by CPCTC. Note the decision that made it work: drawing PS created two triangles that *share* a side (an automatic congruence) and split the base into two congruent halves (a second). When you must add a line to a figure, prefer one that manufactures free facts: midpoints, shared sides, vertical angles. If both base angles measure 55°, the vertex angle is 180° − 2 · 55° = 70°, and the exterior angle at the vertex is 110° — equal to the sum of the two remote interiors, 55° + 55°. The whole lesson in one small triangle.