Lesson 15 ~25 min Unit 3 · Geometry +85 XP

Applying Congruence

Once two triangles are proved congruent, EVERY pair of corresponding sides and EVERY pair of corresponding angles is automatically equal. We'll use this to prove sides equal, angles equal, and to chase missing values.

Today's hook: Prove congruence once. Then SIX equal parts pop out for free.

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Think First

warm-up

An isosceles triangle has two equal sides. Draw the line from the top vertex down to the midpoint of the bottom side — call this the median. Why do you think the two smaller triangles you've created should be the same? What "test" might prove it?

Record your answer in your workbook.

From Congruence to Conclusions

+5 XP

Once you've proved $\triangle ABC \equiv \triangle DEF$, every corresponding piece is equal. That gives you SIX free facts: three pairs of equal sides and three pairs of equal angles. The standard reason phrase is "(matching sides of $\equiv$ $\triangle$s)" or "(matching $\angle$s of $\equiv$ $\triangle$s)".

The strategy is always: first prove the two triangles congruent (using SSS, SAS, AAS or RHS), then use the congruence to deduce the side or angle you want. The vertex order of the congruence statement determines which parts equal which — never read off a fact unless the order matches. Cite the matching reason every time.

Two-step plan: prove $\equiv$, then read off the equal parts.

Always state the test

"(SSS)", "(SAS)", "(AAS)", "(RHS)" — one of these always.

Match the order

$\triangle ABC \equiv \triangle DEF$ means $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$.

Matching reason

"(matching sides/angles of $\equiv$ $\triangle$s)" justifies the next step.

What You'll Master

objectives

Know

The two-step structure: prove $\equiv$, then deduce
How to spot a shared side or shared angle
The matching-parts reason for sides and angles

Understand

Why proving congruence gives SIX equal pieces at once
Why isosceles base angles are equal (a congruence proof)
How to choose the right test from given information

Can Do

Identify a pair of congruent triangles inside a larger figure
Prove two sides equal using congruence
Prove two angles equal using congruence

Words You Need

vocabulary

Matching partsCorresponding sides or angles of two congruent triangles.

Shared sideA side that belongs to BOTH triangles — equal to itself.

Shared angleAn angle at a common vertex — equal to itself.

IsoscelesTriangle with two equal sides; the base angles opposite them are equal.

MedianLine from a vertex to the midpoint of the opposite side.

Perpendicular bisectorA line that cuts a segment in half at $90^{\circ}$.

BisectsCuts exactly in half (equal parts).

"Given"A fact you may use as a starting point in a proof.

Spotting Congruent Triangles in a Figure

+5 XP

In many proofs the two triangles are part of a larger figure. They might share a side (like a diagonal of a quadrilateral) or share an angle (at a common vertex). A shared side or angle is automatically equal to itself — this gives you a "free" piece toward your congruence test.

Common situations:
• A diagonal of a quadrilateral creates two triangles that share that diagonal.
• Two triangles standing on the same base have a common base.
• Two triangles meeting at a vertex have a common angle (vertically opposite angles count too).
Mark the shared piece with a "common" label or with matching tick marks — that single fact often turns "almost congruent" into "definitely congruent".

Shared side / angle $\Rightarrow$ equal to itself $\Rightarrow$ free piece toward the test.

Look for a diagonal

It usually creates two triangles with a shared side.

"Common" is a reason

Write "(common)" or "(shared)" beside the equal piece.

Vertically opposite angles

When triangles meet at a point, look for an "X" with equal angles.

Quick book-notes · Spotting congruent triangles

Look for a shared side, shared angle or shared vertex.
Diagonals of quadrilaterals split them into pairs of triangles.
"Common" or "(shared)" is a valid reason phrase.

Micro-check: In a parallelogram $ABCD$, diagonal $AC$ creates two triangles. Which fact is given for free?

Proving Two Sides Are Equal

+5 XP

To show that two sides of a figure are equal, find two triangles within the figure that contain those sides as corresponding parts, then prove the triangles congruent. The equality of the sides then follows from "matching sides of $\equiv$ $\triangle$s".

Three-step proof template:
1. Identify the two triangles that contain the sides you want to be equal.
2. Prove the triangles congruent (state the three matching pieces and cite the test).
3. Conclude with: "Therefore $XY = X'Y'$ (matching sides of $\equiv$ $\triangle$s)".
This template proves an enormous range of geometric facts — including the classic "isosceles triangle: equal sides give equal base angles" result and its converse.

Matching sides of $\equiv$ $\triangle$s $\Rightarrow$ equal sides.

Find the matching pair

The two sides you want equal must be CORRESPONDING in the congruence statement.

Cite the reason

"(matching sides of $\equiv$ $\triangle$s)" earns the final mark.

Order the vertices

Write the congruence statement carefully so the right pieces line up.

Quick book-notes · Proving sides equal

Step 1: pick the two triangles holding the target sides.
Step 2: prove the triangles congruent.
Step 3: deduce side equality with the matching reason.

True or false: If you prove two triangles congruent, every pair of corresponding sides AND every pair of corresponding angles is automatically equal.

That's exactly the power of congruence — six equal pieces from one proof.

Proving Two Angles Are Equal

+5 XP

The angle-equal version of the proof works the same way: find two triangles with the target angles as corresponding angles, prove them congruent, then cite "matching $\angle$s of $\equiv$ $\triangle$s". This is the key to proving properties of parallelograms, isosceles triangles, kites, and rhombuses.

Example: prove that the diagonals of a rhombus bisect its vertex angles.
1. The two triangles formed by a diagonal share the diagonal (common side), and the rhombus has four equal sides — pick two pairs of sides that match.
2. Triangles congruent by SSS.
3. Therefore the two angles at the vertex where the diagonal meets are equal (matching $\angle$s of $\equiv$ $\triangle$s), i.e. the diagonal bisects the vertex angle. Same template proves diagonals bisect each other in a parallelogram (this time using AAS with alternate angles).

Matching $\angle$s of $\equiv$ $\triangle$s $\Rightarrow$ equal angles.

Same recipe as sides

Two triangles, prove congruent, deduce the angle.

Two angles, one diagonal

Many proofs use a diagonal to split into congruent halves.

Cite matching $\angle$s

"(matching $\angle$s of $\equiv$ $\triangle$s)" is the final-step reason.

Quick book-notes · Proving angles equal

Find two triangles with the target angles as corresponding parts.
Prove the triangles congruent.
Cite "(matching $\angle$s of $\equiv$ $\triangle$s)".

Micro-check: In an isosceles triangle $ABC$ with $AB = AC$, you drop a median $AM$ to the midpoint $M$ of $BC$. Which test proves $\triangle ABM \equiv \triangle ACM$?

Watch Me Solve It · 3 examples

Watch Me Solve It · Isosceles base angles

+15 XP per step

PROBLEM

$\triangle ABC$ has $AB = AC$. $M$ is the midpoint of $BC$. Prove that $\angle B = \angle C$.

1
List the equal parts in $\triangle ABM$ and $\triangle ACM$
$AB = AC$ (given); $BM = CM$ ($M$ midpoint); $AM = AM$ (common)
2
Prove congruence
$\triangle ABM \equiv \triangle ACM$ (SSS)
3
Deduce
$\angle B = \angle C$ (matching $\angle$s of $\equiv$ $\triangle$s)
Classic isosceles result: equal sides $\Rightarrow$ equal base angles.

Answer$\angle B = \angle C$ (matching $\angle$s of $\equiv$ $\triangle$s).

Watch Me Solve It · Diagonal of a parallelogram

+15 XP per step

PROBLEM

$ABCD$ is a parallelogram. The diagonal $AC$ is drawn. Prove $AB = CD$.

1
Find equal parts in $\triangle ABC$ and $\triangle CDA$
$\angle BAC = \angle DCA$ (alternate $\angle$s, $AB \parallel DC$); $\angle BCA = \angle DAC$ (alternate $\angle$s, $AD \parallel BC$); $AC = CA$ (common)
2
Prove congruence
$\triangle ABC \equiv \triangle CDA$ (AAS)
3
Deduce
$AB = CD$ (matching sides of $\equiv$ $\triangle$s)
Same proof shows $BC = DA$ — opposite sides of a parallelogram are equal.

Answer$AB = CD$ (matching sides of $\equiv$ $\triangle$s).

Watch Me Solve It · Find the missing length

+15 XP per step

PROBLEM

$\triangle ABC \equiv \triangle DEF$. $AB = 6.5$ cm, $BC = 8$ cm and $\angle B = 50^{\circ}$. Find $DE$, $EF$ and $\angle E$.

1
Match the vertices
$A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$
2
Read off the matching sides and angle
$DE = AB = 6.5$ cm; $EF = BC = 8$ cm; $\angle E = \angle B = 50^{\circ}$
3
Cite the reasons
All three follow from "(matching sides/$\angle$s of $\equiv$ $\triangle$s)".
Once $\equiv$ is established, ALL corresponding pieces are equal.

Answer$DE = 6.5$ cm, $EF = 8$ cm, $\angle E = 50^{\circ}$.

Common Pitfalls

heads-up

Skipping the congruence proof

Writing "$AB = CD$ because the triangles are congruent" is circular if you haven't proved congruence first.

Fix: Always show the three matching pieces and cite SSS/SAS/AAS/RHS BEFORE deducing.

Reading off the wrong pair

$\triangle ABC \equiv \triangle DEF$ does NOT mean $AB = EF$ — it means $AB = DE$.

Fix: Match vertices in the order of the statement; sides connect matched vertices.

Forgetting the "common" side

A shared side is often the third missing piece in a proof — ignoring it leaves you a piece short.

Fix: Always check if the two triangles share a side or angle.

Copy Into Your Books

Two-step plan

1. Prove triangles congruent
2. Deduce the equal piece
3. Cite the matching reason

Reasons

"(common)" for shared sides/angles
"(matching sides of $\equiv$ $\triangle$s)"
"(matching $\angle$s of $\equiv$ $\triangle$s)"

Common setups

Diagonal of quadrilateral
Isosceles median
Vertically opposite vertex

Watch for

Right order of vertices
Shared / common pieces
Test name in brackets

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Use the Congruence

4 problems

Four quick drills using congruence to find lengths and angles. Solve, then reveal.

1 $\triangle PQR \equiv \triangle XYZ$. $PR = 11$ cm. Find $XZ$.

$P \leftrightarrow X$, $R \leftrightarrow Z$.$XZ = 11$ cm
2 $\triangle ABC \equiv \triangle DEF$. $\angle A = 40^{\circ}$, $\angle B = 75^{\circ}$. Find $\angle F$.

$\angle C = 180 - 40 - 75 = 65^{\circ}$; $\angle F = \angle C$.$\angle F = 65^{\circ}$
3 Isosceles $\triangle ABC$: $AB = AC = 9$ cm. Median $AM$ drawn. Which two triangles are congruent and by which test?

Both have $AM$ common, equal sides, equal halves of $BC$.$\triangle ABM \equiv \triangle ACM$ (SSS)
4 $\triangle KLM \equiv \triangle NOP$ by RHS. $KM = 13$ cm (hyp), $LM = 5$ cm. Find $OP$.

$L \leftrightarrow O$, $M \leftrightarrow P$.$OP = 5$ cm

Complete in your workbook.

Quick Check · 5 questions

When two triangles are proved congruent, how many pairs of corresponding parts are guaranteed equal?

+10 XP

Isosceles $\triangle ABC$ with $AB = AC$; $M$ is midpoint of $BC$. The two sub-triangles are congruent by:

+10 XP

$\triangle ABC \equiv \triangle DEF$. Which is the matching angles statement?

+10 XP

In parallelogram $ABCD$, drawing diagonal $AC$ produces two triangles congruent by:

+10 XP

+10 XP, +5 coins" data-feedback-wrong="Matching parts. Once congruent, sides equal by "matching sides of $\equiv$ $\triangle$s".">

After proving $\triangle ABC \equiv \triangle DEF$, which reason justifies $AB = DE$?

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

Apply Easy 3 MARKS

Q6. $\triangle ABC \equiv \triangle PQR$. $AB = 8$ cm, $\angle A = 55^{\circ}$, $\angle B = 65^{\circ}$.
(a) Find $PQ$ and give the reason.
(b) Find $\angle Q$ and give the reason.
(c) Find $\angle R$ using the triangle angle sum.

Answer in your workbook.

Reason Medium 3 MARKS

Q7. $\triangle ABC$ is isosceles with $AB = AC$. $M$ is the midpoint of $BC$. Prove $\angle ABM = \angle ACM$.
(a) State three equal parts in $\triangle ABM$ and $\triangle ACM$.
(b) Name the congruence test.
(c) State the conclusion with reason.

Answer in your workbook.

Reason Hard 3 MARKS

Q8. In quadrilateral $ABCD$, $AB = CD$ and $AB \parallel CD$. The diagonal $AC$ is drawn.
(a) Show that $\triangle ABC \equiv \triangle CDA$.
(b) Hence prove $BC = AD$.
(c) State which quadrilateral $ABCD$ must be.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. C — 6 pairs (3 sides + 3 angles).

2. B — SSS.

3. A — $\angle A = \angle D$.

4. D — AAS.

5. B — matching sides of $\equiv$ $\triangle$s.

Show Your Working Model Answers

Q6 (3 marks): (a) $PQ = AB = 8$ cm (matching sides of $\equiv$ $\triangle$s) [1]. (b) $\angle Q = \angle B = 65^{\circ}$ (matching $\angle$s of $\equiv$ $\triangle$s) [1]. (c) $\angle R = 180 - 55 - 65 = 60^{\circ}$ ($\angle$ sum of $\triangle$) [1].

Q7 (3 marks): (a) In $\triangle ABM$ and $\triangle ACM$: $AB = AC$ (given), $BM = CM$ ($M$ is midpoint), $AM = AM$ (common) [1]. (b) $\triangle ABM \equiv \triangle ACM$ (SSS) [1]. (c) $\angle ABM = \angle ACM$ (matching $\angle$s of $\equiv$ $\triangle$s) [1].

Q8 (3 marks): (a) In $\triangle ABC$ and $\triangle CDA$: $AB = CD$ (given), $\angle BAC = \angle DCA$ (alternate $\angle$s, $AB \parallel CD$), $AC = CA$ (common). $\therefore \triangle ABC \equiv \triangle CDA$ (SAS) [1]. (b) $BC = DA$ (matching sides of $\equiv$ $\triangle$s) [1]. (c) Both pairs of opposite sides are equal and one pair ($AB \parallel CD$) is parallel; combined with $AB = CD$ this makes $ABCD$ a parallelogram [1].

Stretch Challenge · +25 XP, +10 coins

Diagonals of a Rhombus

$ABCD$ is a rhombus (4 equal sides). Its two diagonals $AC$ and $BD$ meet at point $X$. (a) Prove that $\triangle ABX \equiv \triangle CBX$ using SSS — this shows that $BD$ bisects the vertex angle at $B$. (b) Deduce that $\angle AXB = \angle CXB$, and hence that $AC \perp BD$. (c) State, in one sentence each, two properties of rhombus diagonals you have now proved.

Reveal solution

(a) Easier route: first prove $\triangle ABD \equiv \triangle CBD$ by SSS ($AB = CB$ given, $AD = CD$ given, $BD = BD$ common). This gives $\angle ABD = \angle CBD$ (matching $\angle$s). Then in $\triangle ABX$ and $\triangle CBX$: $AB = CB$, $\angle ABX = \angle CBX$ (just shown), $BX = BX$ (common). $\therefore \triangle ABX \equiv \triangle CBX$ (SAS). (b) From this congruence, $\angle AXB = \angle CXB$ (matching $\angle$s of $\equiv$ $\triangle$s). But $\angle AXB$ and $\angle CXB$ are angles on a straight line at $X$, so they add to $180^{\circ}$. Two equal angles summing to $180^{\circ}$ are each $90^{\circ}$. $\therefore AC \perp BD$. (c) (i) Each diagonal of a rhombus bisects the vertex angles. (ii) The diagonals of a rhombus are perpendicular.

Quick Review

Strategy

Prove $\equiv$, then deduce equal parts.

Reasons

"matching sides/$\angle$s of $\equiv$ $\triangle$s".

Common pieces

Shared sides / angles are equal to themselves.

Isosceles

Equal sides $\Rightarrow$ equal base angles (via SSS).

Parallelogram

Diagonal makes two $\equiv$ triangles (AAS).

Rhombus

Diagonals bisect vertex angles AND are perpendicular.

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