Triangle Congruence Theorems

📐 Triangle Congruence Theorems

Part 1 of 5 — What Congruence Means

Topics in This Part

🔑 Key Concept: Two triangles are congruent when they have exactly the same size and shape — one could be slid, flipped, or rotated to land perfectly on top of the other. Every congruence theorem in this lesson is a shortcut for proving that without checking all six parts.

Congruent vs. Equal

A triangle has six parts: three sides and three angles. If two triangles are congruent, all six corresponding parts match.

• We say the figures are congruent and write $\triangle ABC \cong \triangle DEF$.
• We say the measures are equal and write $AB = DE$ or $m\angle A = m\angle D$.

💡 Wording matters: Segments and angles are congruent ($\cong$); their lengths and degree measures are equal ($=$). On a test, "$\overline{AB} \cong \overline{DE}$" and "$AB = DE$" say the same thing in two correct grammars.

The order of the letters is a promise

The statement $\triangle ABC \cong \triangle DEF$ lists corresponding vertices in order:

Concept Check 🎯

Included Sides and Included Angles

The congruence theorems depend on where the matched parts sit. Two vocabulary words make every theorem precise.

• An included angle is the angle between two named sides (its vertex is where they meet).
• An included side is the side between two named angles (it joins their vertices).

Example: $\triangle ABC$

⚠️ "Included" is the difference between a theorem that works (SAS) and one that doesn't (SSA). Get comfortable spotting it now.

Spot the Included Part 🔽

Use $\triangle XYZ$ for each.

The Payoff: CPCTC

Why prove triangles congruent at all? Because of one powerful follow-up rule:

🔑 CPCTC — Corresponding Parts of Congruent Triangles are Congruent. Once you've proven $\triangle ABC \cong \triangle DEF$, you may declare any pair of matched sides or angles congruent.

That's the whole game plan for most geometry proofs:

1. Use a congruence theorem (SSS, SAS, ASA, AAS, or HL) to prove two triangles congruent.
2. Use CPCTC to conclude that some specific side or angle you actually wanted is congruent.

In Parts 2–4 we'll collect the five theorems. In Part 5 we'll chain them with CPCTC.

Concept Check 🎯

📐 Triangle Congruence Theorems

Part 2 of 5 — SSS and SAS

🔑 The Big Idea: You don't need all six parts. The right three parts lock a triangle's size and shape completely. The first two shortcuts are SSS and SAS.

SSS — Side-Side-Side

🔑 SSS Postulate: If the three sides of one triangle are congruent to the three sides of another, the triangles are congruent.

If $AB = DE$, $BC = EF$, and $CA = FD$, then $\triangle ABC \cong \triangle DEF$.

Three fixed side lengths can build only one triangle (up to flips and turns) — that rigidity is exactly why triangular braces hold up bridges and bicycle frames.

Example

In $\triangle ABC$ and $\triangle DEF$: $AB = 5$, $BC = 7$, $CA = 6$ and $DE = 5$, $EF = 7$, $FD = 6$.

All three pairs of sides match, so $\triangle ABC \cong \triangle DEF$ by SSS.

SAS — Side-Angle-Side

🔑 SAS Postulate: If two sides and the angle between them in one triangle are congruent to the corresponding two sides and included angle of another, the triangles are congruent.

The angle must be the included angle — the one trapped between the two sides.

Example

In $\triangle GHI$ and $\triangle JKL$: $GH = JK = 8$, $\angle H = \angle K = 40°$, and $HI = KL = 5$.

The angle $\angle H$ is included between sides $\overline{GH}$ and $\overline{HI}$ (and $\angle K$ between $\overline{JK}$ and $\overline{KL}$). So $\triangle GHI \cong \triangle JKL$ by SAS.

⚠️ Order check for SAS: the letters read S-A-S — side, included angle, side. If the angle is not between the two sides, this is "SSA," which is not a valid theorem (more on that in Part 3).

Concept Check 🎯

SAS Requires the Included Angle 🔽

For $\triangle ABC$, decide whether each given set is enough for SAS.

Find the Missing Measure 🧮

Because congruent triangles have equal corresponding parts, a known triangle tells you the unknown one.

Given $\triangle ABC \cong \triangle DEF$ with $AB = 12$, $BC = 9$, $CA = 15$, and $m\angle A = 53°$, $m\angle B = 90°$.

1) $DE = \,?$ 2) $EF = \,?$ 3) $m\angle D = \,?$ (in degrees, just the number)

📐 Triangle Congruence Theorems

Part 3 of 5 — ASA, AAS, and the Two Fakes

🔑 The Big Idea: When the matched information includes two angles, the shortcuts are ASA and AAS. Two famous look-alikes — SSA and AAA — are not theorems. Knowing why they fail is as important as knowing the four that work.

ASA — Angle-Side-Angle

🔑 ASA Postulate: If two angles and the side between them (the included side) in one triangle are congruent to the corresponding parts of another, the triangles are congruent.

Example

In $\triangle ABC$ and $\triangle DEF$: $\angle A = \angle D = 50°$, $AB = DE = 7$, and $\angle B = \angle E = 60°$.

Side $\overline{AB}$ is included between $\angle A$ and $\angle B$, so $\triangle ABC \cong \triangle DEF$ by ASA.

💡 Two angles fix the triangle's shape; one matching side fixes its scale. Together they pin down a single triangle.

AAS — Angle-Angle-Side

🔑 AAS Theorem: If two angles and a non-included side of one triangle are congruent to the corresponding parts of another, the triangles are congruent.

Why is AAS automatically valid? Because the Triangle Angle Sum is always $180°$: if two pairs of angles match, the third pair must match too. So a non-included side is actually included between two known angles — AAS quietly becomes ASA.

Example

$\angle A = \angle D = 40°$, $\angle B = \angle E = 75°$, and $BC = EF = 10$. Side $\overline{BC}$ is opposite $\angle A$, not between the two angles — so this is AAS, and the triangles are congruent.

ASA or AAS? 🎯

The Two Fakes: SSA and AAA

⚠️ SSA is NOT a theorem. Two sides and a non-included angle can produce two different triangles — this is the famous "ambiguous case." With one angle and the side opposite it, the third side can sometimes swing to two positions, giving two non-congruent triangles.

⚠️ AAA is NOT a theorem. Three matching angles guarantee the same shape but say nothing about size. Two triangles can have all angles equal yet be different sizes — they are similar, not congruent. (A small $30°$-$60°$-$90°$ triangle and a giant one have identical angles.)

The complete scorecard

Theorem or Not? 🔽

Decide whether each combination guarantees congruence.

Use the Angle Sum 🧮

This is the reasoning behind AAS. The three interior angles of any triangle add to $180°$.

1) A triangle has angles $40°$ and $75°$. The third angle $= \,?$ (degrees) 2) In $\triangle ABC \cong \triangle DEF$, $m\angle A = 90°$ and $m\angle B = 35°$. Then $m\angle F = \,?$ (degrees)

📐 Triangle Congruence Theorems

Part 4 of 5 — HL & Proving with CPCTC

🔑 The Big Idea: Right triangles get a fifth shortcut — HL — and it's the only place a version of "SSA" is allowed. Then we put all five theorems to work inside two-column proofs using CPCTC.

HL — Hypotenuse-Leg (right triangles only)

🔑 HL Theorem: In two right triangles, if the hypotenuse and one leg of one are congruent to the hypotenuse and a leg of the other, the triangles are congruent.

Two conditions must hold to even consider HL:

1. Both triangles are right triangles (there is a $90°$ angle).
2. You match the hypotenuse and one leg.

Why is this allowed when ordinary SSA is not? The right angle plus the Pythagorean Theorem forces the second leg to a single length ($\text{leg}_2 = \sqrt{\text{hyp}^2 - \text{leg}_1^2}$), so the ambiguity disappears and it effectively becomes SSS.

Example

$\triangle ABC$ and $\triangle DEF$ are right triangles with right angles at $C$ and $F$. If hypotenuse $AB =$ hypotenuse $DE = 13$ and leg $BC =$ leg $EF = 5$, then $\triangle ABC \cong \triangle DEF$ by HL. (The third sides are both $\sqrt{13^2 - 5^2} = 12$.)

Does HL Apply? 🎯

Reusing Shared Parts in Proofs

Real proofs rarely hand you all three pairs directly. Two "free" facts close most gaps:

• Reflexive Property: any side or angle is congruent to itself — $\overline{AB} \cong \overline{AB}$. A shared side between two triangles is automatically one matched pair.
• Vertical Angles formed by crossing segments are congruent — often the angle in an SAS or ASA setup.

Worked Proof Sketch

Given: $\overline{AB} \cong \overline{CB}$ and $\overline{DB}$ bisects $\angle ABC$. Prove: $\triangle ABD \cong \triangle CBD$.

The matched parts are S ($AB,CB$), A ($\angle ABD, \angle CBD$), S ($BD,BD$) — and the angle is included between the two sides. ✅

Finish the Proof 🔽

Given: $\overline{AD} \parallel \overline{BC}$ and $\overline{AD} \cong \overline{BC}$, with diagonals crossing so that $A,B$ and $D,C$ are opposite vertices of $\triangle ABD$ and $\triangle CDB$ sharing diagonal $\overline{BD}$. Prove $\triangle ABD \cong \triangle CDB$.

Using CPCTC 🎯

📐 Triangle Congruence Theorems

Part 5 of 5 — Mixed Practice & Mastery Check

You now have all five theorems plus CPCTC. Let's put them together and finish with an Exit Quiz.

Quick Reference

⚠️ Not theorems: SSA (ambiguous — two possible triangles) and AAA (same shape, possibly different size ⇒ only similar).

🔑 Proof game plan: match three correct parts → name the theorem → invoke CPCTC for the side or angle you actually need.

Name That Theorem 🔽

Choose the theorem each set of marks proves (or "none").

Mixed Practice 🎯

Solve for the Unknowns 🧮

$\triangle ABC \cong \triangle PQR$. From the diagram, $AB = 2x + 1$ and its corresponding side $PQ = 11$. Also $m\angle C = 4y$ corresponds to $m\angle R = 72°$.

1) Solve for $x$. (Hint: $2x + 1 = 11$) 2) Solve for $y$. (Hint: $4y = 72$)

Exit Quiz ✅

Answer all three to finish the lesson.