Worksheet on congruent triangles with problems involving identifying congruent triangles and proving triangle congruence using given angles and side lengths.

Let's solve the two problems step by step.

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Problem 1: Which triangles are congruent? Give reasons.

We are given six triangles labeled A through F, each with some angles and side lengths marked. We need to determine which pairs of triangles are congruent, meaning they have exactly the same shape and size (all corresponding sides and angles are equal).

We'll use the triangle congruence criteria:
- ASA – Angle-Side-Angle
- SAS – Side-Angle-Side
- SSS – Side-Side-Side
- AAS – Angle-Angle-Side
- RHS – Right angle-Hypotenuse-Side (only for right triangles)

But note: AAA is not sufficient for congruence (only similarity), and SSA is not valid unless it’s a right triangle.

Let’s analyze each triangle:

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#### Triangle A
- Angles: 60°, 50° → Third angle = 180° - 60° - 50° = 70°
- One side: 8 cm (adjacent to 50° and 60°)
- So angles: 60°, 50°, 70°; side of 8 cm between 60° and 50°

Wait — actually, from the diagram:
- The 8 cm side is opposite the 50° angle?
But we don’t know that directly.

Wait — let’s interpret carefully.

In triangle A:
- Two angles: 60° and 50° → so third angle = 70°
- One side: 8 cm — but which side?

Looking at the diagram: the side of 8 cm is between the 60° and 50° angles? Or opposite one?

But in standard labeling, if no specific vertex is named, we assume the side is adjacent to the angles shown.

Actually, since the 8 cm side is drawn between the 60° and 50° angles, it's the included side.

So:
- Triangle A: 60°, 50°, 70°; side of 8 cm between 60° and 50° → this is ASA or AAS info.

Wait — let's list all known info.

Let’s go one by one.

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#### Triangle A
- Angles: 60°, 50° → third angle = 70°
- Side: 8 cm — appears to be opposite the 50° angle? Wait — let's look.

Actually, in triangle A:
- Vertex labeled A has 60°
- One angle is 50°, and 8 cm side is next to it
- From diagram: the side labeled 8 cm is not between the 60° and 50° angles — wait, it's drawn as one side, and the 50° angle is at the bottom-right, 60° at top-left.

Assuming standard placement:
- The 8 cm side is opposite the 50° angle? No — actually, looking at how it's drawn, the 8 cm side is adjacent to both angles — likely the included side between 60° and 50°?

But we can't be sure without labels.

Alternatively, perhaps we should compute missing angles and compare.

Let’s find all three angles for each triangle.

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Step 1: Find all angles in each triangle

#### Triangle A
- Given: 60°, 50° → third angle = 180 - 60 - 50 = 70°
- Side: 8 cm — position unknown yet

#### Triangle B
- Given: 60°, 50° → third angle = 70°
- Side: 8 cm — again, where is it?

Same as A? Possibly.

But let's see the diagrams:

From the layout:
- Triangle A: 60° at top, 50° at bottom-right, 8 cm on right side.
- Triangle B: 60° at top, 50° at bottom-right, 8 cm on right side — same orientation.

So both A and B have:
- Angles: 60°, 50°, 70°
- Side of 8 cm adjacent to both 60° and 50° → so included side

Thus, A and B satisfy ASA (Angle-Side-Angle): two angles and the included side are equal.

So A ≅ B by ASA

Now check others.

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#### Triangle C
- Angles: 35°, 70° → third angle = 180 - 35 - 70 = 75°
- Side: 6 cm — appears to be opposite the 35° angle?

Wait: in triangle C:
- 35° and 70° shown
- Side of 6 cm is between them? Or adjacent?

Looking at diagram: 6 cm side is between the 35° and 70° angles → so included side

So:
- Angles: 35°, 70°, 75°
- Included side: 6 cm

Compare to others.

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#### Triangle D
- Angles: 40°, and 6 cm, 8 cm sides
- 6 cm and 8 cm are two sides
- Angle between them? Not shown.

But we see:
- One angle is 40°
- Sides: 6 cm and 8 cm — but not clear which side is opposite what.

Wait: the 40° angle is at the bottom-left, and sides are 6 cm (left) and 8 cm (top). So 6 cm and 8 cm are adjacent to the 40° angle.

So: two sides and the included angle?

Yes: sides 6 cm and 8 cm, included angle 40° → SAS?

But we need to know if other triangles match.

Wait: triangle D has:
- 6 cm and 8 cm sides
- Included angle: 40°

So SAS condition.

Now triangle F:
- 8 cm and 6 cm sides
- Angle of 40° — is it included?

Look: triangle F has:
- 8 cm side on left
- 6 cm side on bottom
- 40° angle at bottom-left

So yes — angle between 8 cm and 6 cm → included angle 40°

So triangle D and F both have:
- Two sides: 6 cm and 8 cm
- Included angle: 40°

So D ≅ F by SAS

But wait — triangle E?

#### Triangle E
- Angles: 35°, 70° → third angle = 75°
- Side: 6 cm — is it between the 35° and 70°?

Yes — appears to be the included side.

So triangle E has:
- 35°, 70°, 75°
- Included side: 6 cm

Compare to triangle C:
- 35°, 70°, 75°
- Included side: 6 cm

So C ≅ E by ASA (or AAS — but ASA since side is between the two angles)

So now we have:

- A and B: both 60°, 50°, 70°, 8 cm between 60° and 50° → ASA → A ≅ B
- C and E: both 35°, 70°, 75°, 6 cm between 35° and 70° → ASA → C ≅ E
- D and F: both 6 cm, 8 cm, included angle 40° → SAS → D ≅ F

Are any others congruent?

Check triangle A vs D?

No — different angles.

So only pairs are:
- A ≅ B
- C ≅ E
- D ≅ F

Wait — but is there any overlap?

Let’s double-check.

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#### Recheck triangle D:
- 6 cm, 8 cm, included angle 40° → SAS

Triangle F:
- 8 cm, 6 cm, included angle 40° → same → D ≅ F

Triangle A:
- Angles: 60°, 50°, 70°
- Side: 8 cm — is it between 60° and 50°?

Yes — from diagram: the 8 cm side is between the 60° and 50° angles → so ASA: 60°, 8 cm, 50° → same as B

B has same: 60°, 50°, 8 cm between them → A ≅ B

Triangle C:
- 35°, 70°, 6 cm between them → third angle 75°

Triangle E:
- 35°, 70°, 6 cm between them → same → C ≅ E

So final answer:

> Congruent pairs:
> - A and B → ASA (two angles and included side)
> - C and E → ASA
> - D and F → SAS (two sides and included angle)

Note: No triangle is congruent to more than one other.

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✔ Answer to Problem 1:

- Triangles A and B are congruent by ASA (60°, 8 cm, 50°).
- Triangles C and E are congruent by ASA (35°, 6 cm, 70°).
- Triangles D and F are congruent by SAS (6 cm, 40°, 8 cm).

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Problem 2: Prove that triangles ABC and DEF are congruent.

Given:

- Triangle ABC:
- ∠A = 70°
- ∠C = 50°
- Side BC = 10 cm

- Triangle DEF:
- ∠E = 60°
- ∠F = 70°
- Side DE = 10 cm

Wait — let’s label carefully.

From diagram:

#### Triangle ABC:
- Point A: 70°
- Point C: 50°
- Side BC = 10 cm (between B and C)

So:
- ∠A = 70°
- ∠C = 50°
- Therefore, ∠B = 180° - 70° - 50° = 60°

So angles: A=70°, B=60°, C=50°
Side BC = 10 cm → this is the side opposite angle A

Wait: side BC is opposite angle A → yes.

So:
- In △ABC:
- ∠A = 70°
- ∠B = 60°
- ∠C = 50°
- Side BC = 10 cm → opposite ∠A

#### Triangle DEF:
- Point E: 60°
- Point F: 70°
- Side DE = 10 cm

So:
- ∠E = 60°
- ∠F = 70°
- Then ∠D = 180° - 60° - 70° = 50°

So angles: D=50°, E=60°, F=70°

Side DE = 10 cm → DE is between D and E → so opposite ∠F

Now compare:

△ABC:
- ∠A = 70°, ∠B = 60°, ∠C = 50°
- Side BC = 10 cm → opposite ∠A

△DEF:
- ∠D = 50°, ∠E = 60°, ∠F = 70°
- Side DE = 10 cm → opposite ∠F

Now, notice:

- ∠A = 70° = ∠F
- ∠B = 60° = ∠E
- ∠C = 50° = ∠D

So angles are equal, but order differs.

Now, side BC = 10 cm → opposite ∠A = 70°

Side DE = 10 cm → opposite ∠F = 70°

So:
- In △ABC: side opposite 70° is 10 cm
- In △DEF: side opposite 70° is 10 cm

Also, both triangles have angles: 70°, 60°, 50°

So they are similar — but are they congruent?

Yes — because one pair of corresponding sides is equal, and all angles are equal → so they are congruent by AAS or ASA.

But let’s find corresponding vertices.

Match angles:

- ∠A = 70° = ∠F
- ∠B = 60° = ∠E
- ∠C = 50° = ∠D

So:
- A ↔ F
- B ↔ E
- C ↔ D

Then side BC corresponds to side FE?

Wait: BC connects B and C → corresponds to E and D → ED?

ED is same as DE.

BC = 10 cm → corresponds to ED = 10 cm → yes.

And ∠B = 60° = ∠E, ∠C = 50° = ∠D → so angle-side-angle?

Wait: in △ABC: ∠B = 60°, side BC = 10 cm, ∠C = 50° → so ASA

In △DEF: ∠E = 60°, side ED = 10 cm, ∠D = 50° → same

So:
- ∠B = ∠E = 60°
- BC = ED = 10 cm
- ∠C = ∠D = 50°

So ASA congruence: two angles and included side.

Therefore, △ABC ≅ △DEF by ASA

But note: the correspondence is:
- A ↔ F
- B ↔ E
- C ↔ D

So we write: △ABC ≅ △FED — but usually we write in order.

Since the problem says "prove that triangles ABC and DEF are congruent", we must show that the correspondence works.

But in DEF, the angles are:
- D = 50°
- E = 60°
- F = 70°

So to match ABC:
- A = 70° → matches F
- B = 60° → matches E
- C = 50° → matches D

So ABC ≅ FED

But the problem says “ABC and DEF” — so unless we reorder, they are not directly matching.

But congruence allows relabeling.

So we can say:

△ABC ≅ △FED by ASA, but since the problem asks to prove ABC and DEF are congruent, we need to verify if DEF can be matched.

Wait — triangle DEF has:
- D: 50°
- E: 60°
- F: 70°
- Side DE = 10 cm

So DE is between D and E → angle at D and E → 50° and 60°

In ABC:
- BC is between B and C → angles at B and C: 60° and 50°

So:
- ∠B = 60° = ∠E
- ∠C = 50° = ∠D
- Side BC = DE = 10 cm

So the included side between ∠B and ∠C is BC = 10 cm

The included side between ∠E and ∠D is DE = 10 cm

So by ASA:
- ∠B = ∠E
- BC = DE
- ∠C = ∠D

So △ABC ≅ △EDF — but the problem says DEF.

Wait — DEF means:
- D → first vertex
- E → second
- F → third

But in our matching, we have:
- A ↔ F
- B ↔ E
- C ↔ D

So ABC ↔ FED

But DEF is D-E-F → which would be D↔C, E↔B, F↔A → same as above.

So △ABC ≅ △DEF only if we define the correspondence correctly.

But in standard notation, when we say △ABC ≅ △DEF, it implies:
- A ↔ D
- B ↔ E
- C ↔ F

But here:
- A = 70°, D = 50° → not equal
- So not matching.

Therefore, we cannot say ABC ≅ DEF unless we specify the correspondence.

But the problem says: "Prove that triangles ABC and DEF are congruent."

This implies that with the given labeling, they are congruent — but clearly, from angles:
- ABC: A=70°, B=60°, C=50°
- DEF: D=50°, E=60°, F=70°

So if we map:
- A → F (70°)
- B → E (60°)
- C → D (50°)

Then:
- AB corresponds to FE
- BC corresponds to ED
- CA corresponds to DF

And BC = ED = 10 cm

And angles at B and C match angles at E and D

So △ABC ≅ △FED

But the problem says "DEF" — so perhaps the diagram shows the points differently.

Wait — in triangle DEF:
- E is at top, D at right, F at bottom
- ∠E = 60°, ∠F = 70°, ∠D = 50°
- DE = 10 cm

So DE is from D to E — side between D and E

In ABC:
- BC is from B to C — side between B and C

We already established:
- ∠B = 60° = ∠E
- ∠C = 50° = ∠D
- BC = DE = 10 cm

So △ABC ≅ △EDF by ASA

But the problem wants us to prove ABC ≅ DEF

So unless DEF is meant to be labeled in a way that matches, we need to re-express.

But perhaps the problem assumes the correspondence based on matching angles.

Alternatively, maybe there's a typo, or we misread.

Wait — let’s check side DE = 10 cm

Is DE opposite ∠F? Yes.

In ABC, side BC is opposite ∠A = 70°

In DEF, side DE is opposite ∠F = 70° → so both have a side of 10 cm opposite 70°

So both triangles have:
- An angle of 70°
- An angle of 60°
- An angle of 50°
- And the side opposite the 70° angle is 10 cm

So by AAS or ASA, they are congruent.

But to use AAS:
- Two angles and a non-included side

For example:
- In △ABC: ∠B = 60°, ∠C = 50°, side BC = 10 cm (opposite ∠A)
- But BC is not opposite either 60° or 50° — it's opposite 70°

Wait — BC is opposite ∠A = 70°

So side opposite 70° is 10 cm

Similarly, in △DEF: side opposite 70° (∠F) is DE = 10 cm

So both triangles have:
- ∠A = ∠F = 70°
- ∠B = ∠E = 60°
- ∠C = ∠D = 50°
- Side opposite 70° = 10 cm

So by AAS:
- Two angles and a non-included side (side opposite one of them)

Specifically:
- ∠A = ∠F = 70°
- ∠B = ∠E = 60°
- Side BC = DE = 10 cm → but BC is opposite ∠A, DE is opposite ∠F → so corresponding sides

So AAS applies.

Thus, △ABC ≅ △DEF by AAS, with correspondence:
- A ↔ F
- B ↔ E
- C ↔ D

Even though the labels are different, the triangles are congruent under this mapping.

So the proof is:

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✔ Proof for Problem 2:

In △ABC:
- ∠A = 70°
- ∠C = 50°
- So ∠B = 180° - 70° - 50° = 60°
- Side BC = 10 cm → opposite ∠A

In △DEF:
- ∠E = 60°
- ∠F = 70°
- So ∠D = 180° - 60° - 70° = 50°
- Side DE = 10 cm → opposite ∠F

Now compare:
- ∠A = ∠F = 70°
- ∠B = ∠E = 60°
- Side BC = DE = 10 cm → and both are opposite the 70° angles

So by AAS (Angle-Angle-Side):
- Two angles and a non-included side (opposite one of them) are equal

Therefore, △ABC ≅ △DEF by AAS

(With correspondence: A ↔ F, B ↔ E, C ↔ D)

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✔ Final Answers:

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1. Congruent Triangles:

- A and B are congruent by ASA (60°, 8 cm, 50°)
- C and E are congruent by ASA (35°, 6 cm, 70°)
- D and F are congruent by SAS (6 cm, 40°, 8 cm)

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2. Proof:

- In △ABC: ∠A = 70°, ∠B = 60°, ∠C = 50°, BC = 10 cm
- In △DEF: ∠D = 50°, ∠E = 60°, ∠F = 70°, DE = 10 cm
- So ∠A = ∠F = 70°, ∠B = ∠E = 60°, and side BC = DE = 10 cm (both opposite 70° angles)
- Therefore, by AAS, △ABC ≅ △DEF

✔ Proved.