Triangle Congruence

GE.C.1SSS, SAS, ASA, AAS, HL01/04

Triangle congruence postulates.

Step01of 04

Two triangles are congruent if all six corresponding parts (3 sides, 3 angles) match. But you don't need all six — just the right combination of three pieces.

SSS	→three sides — $A B ≅ D E, B C ≅ E F, A C ≅ D F$
SAS	→two sides AND the angle BETWEEN them
ASA	→two angles AND the side BETWEEN them
AAS	→two angles AND a side NOT between them
HL	→right triangles only — Hypotenuse + one Leg

GE.C.1 — Triangle congruence postulates.

Step 01 of 04

Two triangles are congruent if all six corresponding parts (3 sides, 3 angles) match. But you don't need all six — just the right combination of three pieces.

SSS	→three sides — $A B ≅ D E, B C ≅ E F, A C ≅ D F$
SAS	→two sides AND the angle BETWEEN them
ASA	→two angles AND the side BETWEEN them
AAS	→two angles AND a side NOT between them
HL	→right triangles only — Hypotenuse + one Leg

Step 02 of 04

SAS — the angle must be BETWEEN the two sides. If the angle is somewhere else, you have SSA — which is NOT a congruence postulate (it can produce two different triangles).

SAS: angle is between the two marked sides.

Step 03 of 04

HL is the special right-triangle shortcut. If two right triangles have congruent hypotenuses AND one congruent leg, the triangles are congruent. The right angle (the "A" in SSA) is what saves it from ambiguity.

What does NOT prove congruence:

AAA	→three angles — produces SIMILAR triangles, not congruent
SSA	→two sides + a non-included angle — ambiguous (can give two triangles)

Step 04 of 04

How to prove two triangles congruent.

1. Mark what's given	→tick marks for congruent sides, arcs for congruent angles
2. Look for shared parts	→vertical angles, common sides, parallel lines
3. Pick a postulate	→SSS / SAS / ASA / AAS / HL — match the marks
4. Write the congruence	→ $△ A B C ≅ △ D E F$ in the right order

Key insight

Five postulates cover every congruence proof: SSS, SAS, ASA, AAS, HL. SSA and AAA do NOT work. The order of letters in the congruence statement matters — corresponding parts must line up:

△ A B C ≅ △ D E F

means A↔D, B↔E, C↔F.

GE.C.2 — CPCTC — Corresponding Parts of Congruent Triangles are Congruent.

Step 01 of 04

CPCTC is the most-used theorem in two-column proofs. Once you prove two triangles congruent (using one of the five postulates), CPCTC lets you conclude that EVERY remaining pair of corresponding parts is also congruent.

The order of operations matters: prove the triangles congruent FIRST, then invoke CPCTC.

Step 02 of 04

Worked example. Given: $\overline{A B} ≅ \overline{D C}$ and $\overline{A D} ≅ \overline{B C}$ . Prove: $∠ A ≅ ∠ C$ .

1.	→ $\overline{A B} ≅ \overline{D C}$ (given)
2.	→ $\overline{A D} ≅ \overline{B C}$ (given)
3.	→ $\overline{A C} ≅ \overline{A C}$ (reflexive — shared diagonal)
4.	→ $△ A B C ≅ △ C D A$ (SSS)
5.	→ $∠ A ≅ ∠ C$ (CPCTC)

Step 03 of 04

The reflexive trick. Any segment shared by two triangles is congruent to itself. This sounds trivial but is the workhorse step that makes congruence proofs work — without the shared diagonal in the example above, SSS wouldn't apply.

Same idea for shared angles: $∠ X ≅ ∠ X$ by the reflexive property.

Step 04 of 04

Common CPCTC outcomes worth recognizing.

Bisects an angle	→once you prove triangle congruence, the two halves of the original angle are congruent
Is the midpoint	→two segments to midpoint are congruent
Lines are parallel	→congruent alternate-interior angles, by CPCTC, force parallel lines
Quadrilateral is a parallelogram	→show opposite sides congruent via CPCTC

Key insight

CPCTC is always a "step 2" tool. First prove the triangles congruent, THEN claim the corresponding parts are congruent. Reaching for CPCTC before triangle congruence is established is the most common proof mistake.

GE.C.3 — Triangle angle-sum theorem and exterior angle theorem.

Step 01 of 04

Triangle angle-sum theorem. The three interior angles of any triangle sum to $180°$ . Always. No exceptions.

∠ A + ∠ B + ∠ C = 180°

Consequence: if you know any two angles, the third is $180°$ minus their sum.

Step 02 of 04

Worked example. Two angles of a triangle measure $52°$ and $73°$ . Find the third.

∠3 = 180° - 52° - 73° = 55°

Special cases: in an equilateral triangle, all three angles are $60°$ . In a right triangle, the two non-right angles sum to $90°$ (so they're complementary).

Step 03 of 04

Exterior angle theorem. Extend one side of a triangle past a vertex. The angle formed OUTSIDE equals the sum of the two NON-adjacent interior angles.

Exterior angle = sum of the two remote interior angles.

exterior at B = ∠ A + ∠ C

Step 04 of 04

Why the exterior angle theorem is a shortcut.

$ext + ∠ B = 180°$ (linear pair) and $∠ A + ∠ B + ∠ C = 180°$ (angle sum). Subtract: $ext = ∠ A + ∠ C$ . Saves a step every time you'd otherwise compute the interior at B and then take the supplement.

Key insight

Interior angles always sum to

180°

. Each exterior angle equals the sum of the OTHER two interior angles — not the adjacent one. Both follow from the same angle-sum fact.

GE.C.4 — Triangle inequality.

Step 01 of 04

Triangle inequality theorem. The sum of ANY two sides of a triangle is greater than the third side.

a + b > c, a + c > b, b + c > a

Geometric reason: the shortest path between two points is a straight line, so going "the long way" through a third vertex must be strictly longer.

Step 02 of 04

Quick test. Can sides 3, 5, 9 form a triangle? Check the largest side against the sum of the other two:

3 + 5 = 8 < 9 \Rightarrow NO triangle

Sides 5, 7, 11? $5 + 7 = 12 > 11$ — yes, triangle. (You only need to check the biggest side against the sum of the other two; if THAT passes, the other two inequalities pass automatically.)

Step 03 of 04

Range of possible third sides. Given two sides $a$ and $b$ (with $a \geq b$ ), the third side $c$ must satisfy:

a - b < c < a + b

Lower bound: triangle inequality the other way. Upper bound: triangle inequality applied to the missing side.

Example: sides 7 and 4. Third side must satisfy $3 < c < 11$ — strictly between, not equal.

Step 04 of 04

Side-angle relationship in triangles. The longest side is opposite the largest angle, the shortest side is opposite the smallest angle.

If a > b > c (sides)	→then ∠A > ∠B > ∠C (opposite angles)
If ∠A > ∠B > ∠C	→then a > b > c
Equilateral	→all sides equal → all angles equal (= 60°)
Isosceles	→two sides equal → two opposite angles equal

Key insight

Triangle inequality: sum of any two sides > third side. Equivalent shortcut: the third side lies strictly between the difference and the sum of the other two. Pair this with "longest side opposite largest angle" to handle most range-of-side and ordering questions.

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