Congruent Triangles Worksheets - Math Monks - Free Printable

Let's solve each problem step-by-step from the Congruent Triangles Worksheet.

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Problem 1: Prove whether the given triangles are congruent or not. If yes, state the theorem.

Given:
- Triangle ABC with sides:
- AB = 2 yd
- BC = 3.5 yd
- AC = 4.5 yd
- Triangle PQR with sides:
- PQ = 2 yd
- QR = 4.5 yd
- PR = 3.5 yd

Step-by-step:

Compare the side lengths:

| Triangle ABC | Triangle PQR |
|-------------|--------------|
| AB = 2 yd | PQ = 2 yd |
| BC = 3.5 yd | QR = 4.5 yd → Not matching |
| AC = 4.5 yd | PR = 3.5 yd → Not matching |

Wait! Let’s double-check:

- AB = PQ = 2 yd ✔
- BC = 3.5 yd, but QR = 4.5 yd ✘
- AC = 4.5 yd, but PR = 3.5 yd ✘

But look at the order of vertices:

Triangle ABC has:
- AB = 2 yd
- BC = 3.5 yd
- AC = 4.5 yd

Triangle PQR has:
- PQ = 2 yd
- QR = 4.5 yd
- PR = 3.5 yd

So:
- AB = PQ = 2 yd ✔
- BC = 3.5 yd, PR = 3.5 yd → But PR is not corresponding to BC unless we match properly.
- AC = 4.5 yd, QR = 4.5 yd ✔

Now, check if all three sides match in corresponding order:

Let’s try mapping:
- A ↔ P (since AB and PQ both 2 yd)
- B ↔ Q
- C ↔ R

Then:
- AB = PQ = 2 yd ✔
- BC = QR? BC = 3.5 yd, QR = 4.5 yd ✘
Not equal.

Try another correspondence:
Suppose:
- A ↔ P
- B ↔ R
- C ↔ Q

Then:
- AB = PR? AB = 2 yd, PR = 3.5 yd ✘

Try:
- A ↔ Q
- B ↔ P
- C ↔ R?

This is messy.

Instead, list all sides:

ABC: 2, 3.5, 4.5
PQR: 2, 3.5, 4.5 → same side lengths!

So, the side lengths are identical, just possibly in different order.

Therefore, by SSS (Side-Side-Side) Congruence Theorem, if all three sides of one triangle are equal to the corresponding sides of another, then the triangles are congruent.

We need to find correct correspondence.

From:
- AB = 2 yd → must correspond to a side of length 2 yd in PQR → PQ = 2 yd → So A↔P, B↔Q
- BC = 3.5 yd → must go to a side of 3.5 yd in PQR → PR = 3.5 yd → So B↔P, C↔R → Conflict.

Wait — better to list:

In △ABC:
- AB = 2
- BC = 3.5
- AC = 4.5

In △PQR:
- PQ = 2
- QR = 4.5
- PR = 3.5

So:
- AB = PQ = 2 → A ↔ P, B ↔ Q
- AC = 4.5 → QR = 4.5 → So C ↔ R
- Then BC = 3.5 → PR = 3.5 → B ↔ P, C ↔ R → B ↔ P? But earlier B ↔ Q → conflict.

Better way: Match sides:

- AB = PQ = 2 → A ↔ P, B ↔ Q
- AC = QR = 4.5 → A ↔ Q? No — inconsistent.

Try matching by side lengths:

Both triangles have sides: 2, 3.5, 4.5

So they are congruent by SSS.

The correspondence is:
- AB = PQ = 2 yd → A ↔ P, B ↔ Q
- AC = PR = 3.5 yd? No, AC = 4.5 yd

Wait — AC = 4.5 yd, PR = 3.5 yd → no.

Let’s recheck:

△ABC:
- AB = 2 yd
- BC = 3.5 yd
- AC = 4.5 yd

△PQR:
- PQ = 2 yd
- QR = 4.5 yd
- PR = 3.5 yd

So:
- AB = PQ = 2 yd → A ↔ P, B ↔ Q
- BC = 3.5 yd → PR = 3.5 yd → So C ↔ R
- Then AC = 4.5 yd → QR = 4.5 yd → C ↔ R, A ↔ Q? But A already mapped to P → contradiction.

Wait — so let’s suppose:

A ↔ Q
B ↔ P
C ↔ R

Then:
- AB = QP = 2 yd → yes (QP = PQ = 2 yd) ✔
- BC = PR = 3.5 yd → B→P, C→R → PR = 3.5 yd → yes ✔
- AC = QR = 4.5 yd → A→Q, C→R → QR = 4.5 yd → yes ✔

So correspondence is:
- A ↔ Q
- B ↔ P
- C ↔ R

So △ABC ≅ △QPR

Thus, yes, the triangles are congruent by SSS (Side-Side-Side) Congruence Theorem.

✔ Answer: Yes, △ABC ≅ △QPR by SSS.

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Problem 2: In the given figure, prove that △AEB ≅ △AEC

Given Figure:
- Triangle ABC with AE perpendicular to BC (right angle at E)
- AB = 7 in
- AC = 7 in
- AE is common
- ∠AEB = ∠AEC = 90°

So:
- AB = AC = 7 in → Isosceles triangle
- AE ⊥ BC → AE is altitude
- So BE = EC? Only if it's also median, which happens in isosceles triangle.

Since AB = AC, and AE is altitude from A to BC, then in an isosceles triangle, the altitude to the base is also the median and angle bisector.

So:
- BE = EC
- AE is common
- ∠AEB = ∠AEC = 90°

So in △AEB and △AEC:

- AE = AE (common)
- ∠AEB = ∠AEC = 90°
- BE = EC (from symmetry)

So by SAS (Side-Angle-Side):
- Two sides and included angle equal → SAS

Or even better, since both are right triangles, and hypotenuse and leg are equal:

AB = AC (hypotenuse), AE = AE (leg), so by HL (Hypotenuse-Leg) Congruence Theorem for right triangles.

✔ Answer: △AEB ≅ △AEC by HL (or SAS).

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Problem 3: Given AB ≅ EF and BC ≅ DF, show that △ABD ≅ △EFC

Given:
- AB ≅ EF (given)
- BC ≅ DF (given)
- Right angles at B and F
- BD and FC are parts of line segments

From diagram:
- AB ⊥ BD → ∠B = 90°
- EF ⊥ FC → ∠F = 90°
- Also, BC = CD? Wait — markings: BC and CD have single tick marks → BC = CD
- Similarly, DF has two ticks, and CF? Wait — let's interpret:

Markings:
- AB and EF: both have one tick → AB ≅ EF (given)
- BC and DF: both have one tick → BC ≅ DF (given)
- BD and FC: both have two ticks → BD ≅ FC
- Also, ∠B = ∠F = 90°

Now, we want to prove △ABD ≅ △EFC

List the parts:

In △ABD:
- AB = side
- BD = side
- ∠B = 90°

In △EFC:
- EF = side
- FC = side
- ∠F = 90°

We are given:
- AB ≅ EF (given)
- BD ≅ FC (from markings)
- ∠B ≅ ∠F = 90°

So:
- Two sides and included angle equal → SAS

Thus, △ABD ≅ △EFC by SAS

✔ Answer: △ABD ≅ △EFC by SAS

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Problem 4: Given △DAB ≅ △BCD, find x

Figure:
- Parallelogram ABCD (since opposite sides appear parallel)
- Diagonal DB drawn
- Angles: ∠ADB = 2x² + 7, ∠CBD = 57°
- Given: △DAB ≅ △BCD

From congruence: △DAB ≅ △BCD

So corresponding parts are equal.

Let’s determine correspondence.

Vertices:
- DAB ≅ BCD → D ↔ B, A ↔ C, B ↔ D

So:
- DA ↔ BC
- AB ↔ CD
- DB ↔ BD (common)

Now, angles:
- ∠DAB ↔ ∠BCD
- ∠ABD ↔ ∠CDB
- ∠ADB ↔ ∠CBD

So ∠ADB = ∠CBD

But ∠ADB = 2x² + 7
∠CBD = 57°

So:
2x² + 7 = 57
2x² = 50
x² = 25
x = ±5

But since x represents a measure (likely positive), x = 5

✔ Answer: x = 5

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Problem 5: In the given congruent triangles under ASA, find x and y. Given △ABC ≅ △XYZ

Given:
- △ABC ≅ △XYZ (by ASA)
- In △ABC:
- ∠B = 60°
- ∠C = 30°
- Side BC = 6 cm
- In △XYZ:
- ∠Y = 60°
- Side YZ = 6 cm
- ∠X = x, ∠Z = y

From congruence: △ABC ≅ △XYZ

So correspondence:
- A ↔ X
- B ↔ Y
- C ↔ Z

So:
- ∠A ↔ ∠X
- ∠B ↔ ∠Y = 60° → matches
- ∠C ↔ ∠Z = 30° → so y = 30°
- Side BC ↔ YZ → BC = YZ = 6 cm → matches

Now, sum of angles in triangle = 180°

In △ABC:
- ∠B = 60°, ∠C = 30° → ∠A = 180 - 60 - 30 = 90°

So ∠A = 90° → corresponds to ∠X → x = 90°

Also, ∠Z = ∠C = 30° → y = 30°

✔ Answer: x = 90°, y = 30°

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Problem 6: Complete the congruence statement

#### a) Triangle ABC with:
- AB = AC (two ticks on AB and AC)
- BC = CD (BC and CD have one tick)
- AD is drawn from A to D, and it's perpendicular to BC at C → ∠ACB = 90°
- So △ABC is isosceles with AB = AC, and AD is altitude?

Wait — actually, from diagram:
- AB = AC (ticks)
- BC = CD (ticks)
- AD is drawn, and ∠ACB is right angle?
Wait — the right angle is at C between BC and CD? Or is it at C in triangle ABC?

Looking at diagram:
- Point C is where AB, AC, and CD meet
- Right angle at C between AC and BC? But marking shows right angle symbol at C between AC and CD?

Actually, it looks like:
- AB = AC (equal sides)
- BC = CD (equal segments)
- AD is a line from A to D
- ∠ACB = 90°? But no — the right angle is at C between AC and CD?

Wait — the right angle is marked at point C, between AC and CD → so ∠ACD = 90°?

But we have triangle ABC and triangle ACD?

Wait — the triangle is ABC and ADC?

Wait — perhaps it's △ABC and △ADC?

But the question says: "Complete the congruence statement" for △ABC ≅ ___

From markings:
- AB = AC → AB and AC have two ticks → so AB = AC
- BC = CD → BC and CD have one tick → BC = CD
- AC is common

Wait — this suggests that AC is common side.

But we need to see if there's a triangle congruent to ABC.

Wait — likely, it's showing △ABC ≅ △ACD?

But points: A, B, C and A, C, D

But AB = AC → AB = AC → so AB = AC → but AC is a side, so AB = AC → triangle ABC has AB = AC → isosceles

And BC = CD → so BC = CD

And AC = AC (common)

But we don't know about angles.

Wait — the right angle is at C → ∠ACB = 90°? But it's shown at C between AC and CD → so ∠ACD = 90°?

Wait — perhaps the figure shows:
- Triangle ABC with AB = AC
- Point D such that CD = BC
- And AD is drawn, and ∠ACD = 90°?

But without clear labeling, let's assume standard interpretation.

Alternatively, maybe it's showing two right triangles sharing AC.

But more likely, based on common problems:

It's likely that:
- AB = AC (given)
- BC = CD (given)
- AC = AC (common)
- But angles?

Wait — the right angle is at C → so ∠ACB = 90°? But if AB = AC, and ∠C = 90°, then triangle ABC has AB = AC and ∠C = 90° → impossible because in a triangle, equal sides imply equal angles.

If AB = AC, then ∠B = ∠C

But if ∠C = 90°, then ∠B = 90° → sum > 180° → impossible.

So contradiction.

Ah — perhaps the right angle is at C between AC and CD → so ∠ACD = 90°

Then we have:
- AB = AC (ticks)
- BC = CD (ticks)
- AC = AC (common)

But still not enough.

Wait — perhaps it's intended to be:

△ABC and △ADC?

But AB = AC, BC = CD, AC = AC → SSS?

But AB = AC, so AB = AC → but in △ADC, AC is side, but AD is unknown.

Wait — maybe the triangle is △ABC and △ACD?

But the figure shows point D on extension of BC?

Alternatively, perhaps it's a kite shape.

But looking at the second part:

b) △QRS ≅ ___

With figure: two triangles, PQT and QRS

Points: P, Q, T, S, R

Markings:
- PQ = QT (one tick)
- QS = SR (one tick)
- PT = RS? Wait — PT and RS have two ticks
- Also, ∠P = ∠R = 90°
- ∠T = ∠S (both have arcs)

So:
- PQ = QT → isosceles triangle PQT
- QS = SR → isosceles triangle QSR
- PT = RS (two ticks)
- ∠P = ∠R = 90°
- ∠T = ∠S

But we are to complete: △QRS ≅ ___

From diagram:
- Triangle QRS and triangle PQT?

Wait — triangle QRS and triangle PQT?

But QRS has points Q, R, S

PQT has points P, Q, T

Markings:
- PQ = QT → PQ = QT
- QS = SR → QS = SR
- PT = RS → PT = RS

But PT is in △PQT, RS is in △QRS

Also, ∠P = ∠R = 90°

And ∠T = ∠S

So:
- ∠P = ∠R = 90°
- ∠T = ∠S
- PT = RS (given)

So by AAS (Angle-Angle-Side), △PQT ≅ △QRS

So △QRS ≅ △PQT

But order matters.

Correspondence:
- ∠P ↔ ∠R
- ∠T ↔ ∠S
- PT ↔ RS

So vertex P ↔ R, T ↔ S, Q ↔ Q? No — Q is common?

Wait — Q is shared?

No — Q is in both, but in △QRS and △PQT

So likely:
- P ↔ R
- Q ↔ Q
- T ↔ S

But then PQ ↔ RQ? But PQ and RQ not necessarily equal.

Wait — markings:

- PQ = QT → so PQ = QT
- QS = SR → QS = SR
- PT = RS → PT = RS

So:
- PT = RS (given)
- ∠P = ∠R = 90°
- ∠T = ∠S

So by AAS: △PQT ≅ △QRS

So correspondence:
- P ↔ R
- Q ↔ Q
- T ↔ S

So △PQT ≅ △QRS

Therefore, △QRS ≅ △PQT

✔ Answer: △QRS ≅ △PQT

Now back to a):

a) △ABC ≅ ____

From diagram:
- AB = AC (two ticks)
- BC = CD (one tick)
- AC = AC (common)
- Right angle at C between AC and CD → ∠ACD = 90°
- But ∠ACB is not necessarily 90°

Wait — perhaps the right angle is at C in triangle ABC?

But then AB = AC, ∠C = 90° → impossible as before.

Unless it's not ∠ACB.

Wait — maybe the triangle is △ABC and △ADC?

But no — likely, the figure shows that:

- AB = AC
- BC = CD
- AC = AC
- ∠ACB = ∠ACD = 90°? But only one right angle shown.

Wait — the right angle is at C, between AC and BC? Or AC and CD?

Looking at the diagram: likely, the right angle is at C, and AC is common, and BC = CD, AB = AC, and ∠ACB = ∠ACD = 90°?

But if both angles at C are 90°, then BC and CD are perpendicular to AC, so B and D are on opposite sides.

Then:
- AB = AC
- AC = AC
- BC = CD
- ∠ACB = ∠ACD = 90°

So △ABC and △ACD

But AB = AC, but in △ACD, AC is side, but AD is unknown.

Wait — but AB = AC → so AB = AC → so AB = AC

But in △ACD, we have AC, CD, AD

We don’t have AB = AD.

But if AB = AC and AC = AC, and BC = CD, and ∠ACB = ∠ACD = 90°, then:

In △ABC and △ACD:
- AC = AC (common)
- BC = CD (given)
- ∠ACB = ∠ACD = 90°

So by SAS: △ABC ≅ △ACD

But AB = AC → so AB = AC, but in △ACD, AC is side, but AB is not in △ACD.

Wait — unless AB = AD?

But no info.

Wait — perhaps the congruence is △ABC ≅ △ACD?

But AB and AD are not necessarily equal.

But if AB = AC, and AC = AC, and BC = CD, and ∠ACB = ∠ACD = 90°, then:

- AC = AC
- BC = CD
- ∠ACB = ∠ACD = 90°

So SAS applies → △ABC ≅ △ACD

But the answer should be △ABC ≅ △ACD

But AB = AC → so AB = AC → but in △ACD, AC is a side, so AB = AC → but AB is not a side in △ACD.

But the congruence is based on:
- AC = AC
- BC = CD
- ∠ACB = ∠ACD

So yes, SAS → △ABC ≅ △ACD

But the naming: △ABC and △ACD → so correspondence:
- A ↔ A
- B ↔ C
- C ↔ D

So △ABC ≅ △ACD

But is that valid?

Wait — in △ABC, sides: AB, BC, AC

In △ACD: AC, CD, AD

We have:
- AC = AC
- BC = CD
- ∠ACB = ∠ACD

Yes → SAS → congruent

So △ABC ≅ △ACD

But AB = AC → so AB = AC → but in △ACD, AC is side, so AB = AC → but AB is not a side of △ACD.

But the congruence doesn't require that.

So yes.

But wait — the triangle is labeled ABC and the other is probably ADC.

But the blank is: △ABC ≅ ______

So likely: △ACD

But let’s confirm with markings.

Alternatively, perhaps it’s △ABC ≅ △ADC?

Same thing.

✔ Answer: △ABC ≅ △ACD

But some might write △ADC.

But based on order: A→A, B→C, C→D → so △ABC ≅ △ACD

Yes.

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✔ Final Answers:

1. Yes, △ABC ≅ △QPR by SSS
2. △AEB ≅ △AEC by HL (or SAS)
3. △ABD ≅ △EFC by SAS
4. x = 5
5. x = 90°, y = 30°
6. a) △ABC ≅ △ACD
b) △QRS ≅ △PQT

Let me know if you'd like diagrams or further explanation!