Geometry Worksheet Congruent Triangles | PDF - Free Printable

Let's go through each of the 8 problems on the "Proving Congruent Triangles Worksheet" and determine whether the given pairs of triangles are congruent, and if so, identify the appropriate congruence theorem (SSS, SAS, ASA, AAS, or HL).

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🔷 Problem 1

Triangles: ΔABC and ΔADC
- AC is common (shared side).
- AB = AD (marked with one tick).
- ∠ACB and ∠ACD are both right angles (right angle symbol at C).
- So, we have:
- Hypotenuse: AB = AD
- Leg: AC is common
- Right angles at C

✔ Congruent by HL Theorem (Hypotenuse-Leg for right triangles)

> ✔ Answer: Yes, congruent by HL

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🔷 Problem 2

Triangles: ΔABD and ΔCBD
- BD is common.
- AB = CB (both marked with one tick)
- AD = CD (both marked with two ticks)
- All three sides are equal.

✔ Congruent by SSS (Side-Side-Side)

> ✔ Answer: Yes, congruent by SSS

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🔷 Problem 3

Quadrilateral DEFG, diagonal EG divides it into ΔDEG and ΔFEG
- DE = FG (marked with two ticks)
- DG = FE (marked with one tick)
- EG is common

Wait — let’s look carefully:

Actually, from diagram:
- DE = GF (two ticks)
- DG = EF (one tick)
- EG is shared

But note: the markings show:
- DE = GF
- DG = EF
- EG = GE (common)

So all three sides are equal?

Wait — actually, the triangle pair is ΔDEG and ΔFGE?

But the figure shows a parallelogram-like shape. Let’s check:

- DE = GF (equal sides)
- DG = EF (equal sides)
- EG is common

So yes: ΔDEG ≅ ΔFGE by SSS

But wait — are they labeled correctly?

Looking again: Points D, E, F, G — likely quadrilateral DEFG with diagonal EG.

Then:
- Triangle 1: ΔDEG
- Triangle 2: ΔFEG

Now:
- DE = FG? No — DE has two ticks, FG has two ticks → yes, DE = FG
- DG = FE? DG has one tick, FE has one tick → yes
- EG = GE → common

So: SSS applies.

✔ Answer: Yes, congruent by SSS

> ✔ Answer: Yes, congruent by SSS

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🔷 Problem 4

Triangles: ΔTGU and ΔEGL
- Both have right angles at T and L.
- TU = EL (both marked with one tick)
- TG = EL? Wait — no.

Wait: Markings:
- TU = EL (one tick)
- GU = GL (two ticks)
- And right angles at T and L.

So:
- One leg: TU = EL
- Another leg: GU = GL
- Right angles

So both are right triangles with two legs equal.

✔ Congruent by SAS? Or better — LL (Leg-Leg), which is a case of SAS in right triangles.

But standard theorem: SAS or LL (special case for right triangles).

But since both are right triangles and legs are equal → LL (or SAS)

Alternatively, we can say SAS: two legs and included right angle.

✔ Answer: Yes, congruent by SAS (or LL)

But usually written as SAS

> ✔ Answer: Yes, congruent by SAS

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🔷 Problem 5

Triangles: ΔGIH and ΔKJI
- GI = KJ (one tick)
- HI = JI (two ticks)
- Angle at I: vertical angles? They intersect at point I.

So:
- GI = KJ
- HI = JI
- ∠GIH = ∠KJI (vertical angles) → equal

But are these corresponding?

Wait: ΔGIH and ΔKJI

Vertices:
- G → K?
- I → J?
- H → I?

Wait — perhaps better to label:

From diagram:
- GI = KJ (one tick)
- HI = JI (two ticks)
- ∠GIH and ∠KJI are vertical angles → equal

So we have:
- Side GI = KJ
- Included angle ∠GIH = ∠KJI
- Side HI = JI

Wait — but this would be SAS only if the sides are adjacent to the angle.

But GI and HI are adjacent to ∠I in ΔGIH
KJ and JI are adjacent to ∠J in ΔKJI

And ∠I and ∠J are vertical angles → equal

So yes: SAS

But note: GI = KJ, HI = JI, and ∠I = ∠J → SAS

✔ Answer: Yes, congruent by SAS

> ✔ Answer: Yes, congruent by SAS

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🔷 Problem 6

Triangles: ΔABC and ΔEDC
- Point C is common
- BC = DC (one tick)
- AC = EC (two ticks)
- ∠ACB = ∠ECD (vertically opposite angles → equal)

So:
- Two sides and included angle?

Wait:
- BC = DC
- AC = EC
- ∠ACB = ∠ECD (included between AC & BC, and EC & DC)

Yes! So SAS

✔ Answer: Yes, congruent by SAS

> ✔ Answer: Yes, congruent by SAS

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🔷 Problem 7

Triangles: ΔQRS and ΔTUS
- QS = TS (one tick)
- RS = US (two ticks)
- ∠QSR = ∠TSU (vertically opposite angles → equal)

So:
- Side QS = TS
- Side RS = US
- Included angle ∠QSR = ∠TSU

→ SAS

✔ Answer: Yes, congruent by SAS

> ✔ Answer: Yes, congruent by SAS

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🔷 Problem 8

Parallelogram UVXY, diagonal UY connects U to Y
Triangles: ΔUXY and ΔVYU

Wait — labeling: points U, V, X, Y — likely parallelogram UVXY

Diagonal UY divides it into ΔUXY and ΔVYU?

Wait — better: ΔUXY and ΔVYU?

Wait — maybe ΔUXY and ΔVYU?

Let’s see:

- UX = VY (opposite sides of parallelogram — assumed from markings?)
- XY = VU (opposite sides)
- UY = YU (common diagonal)

Markings:
- UX = VY (one tick)
- XY = VU (one tick)
- UY = YU (common)

Also, from diagram:
- UX = VY (one tick)
- XY = VU (one tick)
- UY is common

So all three sides equal → SSS

But also: opposite sides of parallelogram are equal → so yes.

✔ Answer: Yes, congruent by SSS

> ✔ Answer: Yes, congruent by SSS

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✔ Final Answers Summary:

| Problem | Congruent? | Reason |
|--------|------------|--------|
| 1 | Yes | HL (Right triangle, hypotenuse and leg) |
| 2 | Yes | SSS (Three sides equal) |
| 3 | Yes | SSS (All three sides equal) |
| 4 | Yes | SAS (Two legs and included right angle) |
| 5 | Yes | SAS (Two sides and included angle) |
| 6 | Yes | SAS (Two sides and included angle) |
| 7 | Yes | SAS (Two sides and included angle) |
| 8 | Yes | SSS (All three sides equal) |

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📝 Notes:

- HL is used only for right triangles when hypotenuse and one leg are equal.
- SAS: Two sides and the included angle.
- SSS: Three sides equal.
- Vertical angles are always equal.
- Common sides are shared.

All eight pairs are congruent based on the markings and geometric properties.

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✔ Final Answer Sheet:

1. Yes, HL
2. Yes, SSS
3. Yes, SSS
4. Yes, SAS
5. Yes, SAS
6. Yes, SAS
7. Yes, SAS
8. Yes, SSS

Let me know if you'd like this formatted as a printable answer key!