Geometry worksheet featuring three practice problems for proving congruent triangles using two-column proofs.

Let’s go step by step through each proof on the worksheet. We’ll fill in the “Reasons” column for each statement, based on standard geometry rules and the given information.

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First Proof:

Given:
- $\overline{JG} \cong \overline{HI}$
- $\angle JGI \cong \angle HIG$
Prove: $\angle J \cong \angle H$

We are proving two triangles congruent to then say their corresponding angles are congruent.

Step-by-step:

1. $\overline{JG} \cong \overline{HI}$ → This is given.
2. $\angle JGI \cong \angle HIG$ → Also given.
3. $\overline{GI} \cong \overline{IG}$ → This is the same segment, just written backwards. So it’s the Reflexive Property of Congruence.
4. $\triangle JGI \cong \triangle HIG$ → We have two sides and the included angle? Wait — let’s check:

Actually, we have:
- Side: $\overline{JG} \cong \overline{HI}$
- Angle: $\angle JGI \cong \angle HIG$ (this is between sides JG and GI, and HI and IG)
- Side: $\overline{GI} \cong \overline{IG}$

So yes — Side-Angle-Side → SAS Congruence Postulate

5. $\angle J \cong \angle H$ → These are corresponding parts of congruent triangles → CPCTC (Corresponding Parts of Congruent Triangles are Congruent)

✔ First proof reasons filled.

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Second Proof:

Given:
- $\angle P \cong \angle T$
- R is the midpoint of $\overline{QS}$
Prove: $\triangle PRQ \cong \triangle TRS$

Step-by-step:

1. $\angle P \cong \angle T$ → Given
2. R is the midpoint of $\overline{QS}$ → Given
3. $\overline{QR} \cong \overline{SR}$ → Because R is the midpoint → Definition of Midpoint
4. $\angle PRQ \cong \angle TRS$ → These are vertical angles (formed by intersecting lines PT and QS at point R) → Vertical Angles Theorem
5. $\triangle PRQ \cong \triangle TRS$ → We have:
- $\angle P \cong \angle T$ (angle)
- $\overline{QR} \cong \overline{SR}$ (side)
- $\angle PRQ \cong \angle TRS$ (angle)

Wait — that’s Angle-Side-Angle? But note: the side is between the two angles? Let’s see:

In triangle PRQ: angles at P and R, with side QR between them? Actually, no — side QR is adjacent to angle R and opposite angle P.

Actually, we have:
- Angle at P ≅ angle at T
- Side QR ≅ SR (but this is not between the angles in the same way)
Wait — better to think: we have two angles and a non-included side? Or maybe AAS?

Actually, since we have:
- $\angle P \cong \angle T$
- $\angle PRQ \cong \angle TRS$ (vertical angles)
- and side $\overline{QR} \cong \overline{SR}$ — which is opposite one of the angles? Hmm.

Actually, in triangle PRQ and TRS:
- $\angle P \cong \angle T$
- $\angle PRQ \cong \angle TRS$
- and side $\overline{QR} \cong \overline{SR}$ — but QR is in triangle PRQ, SR is in triangle TRS.

Note: In triangle PRQ, side QR is between angles P and PRQ? No — actually, side QR is opposite angle P.

Wait — perhaps it’s AAS: two angles and a non-included side.

Yes! We have:
- Two angles: $\angle P \cong \angle T$, $\angle PRQ \cong \angle TRS$
- And a side that is NOT between them: $\overline{QR} \cong \overline{SR}$ — which is opposite $\angle P$ and $\angle T$ respectively.

So → AAS Congruence Theorem

Alternatively, some might argue ASA if they consider the side between the angles — but here, the side QR is not between angles P and PRQ — actually, in triangle PRQ, side QR is between angles Q and R, not P and R.

Wait — let’s label properly:

In triangle PRQ:
- Vertices: P, R, Q
- Angles: at P, at R, at Q
- Side QR is between angles at Q and R.

But we don’t know anything about angle Q.

We know:
- Angle at P ≅ angle at T
- Angle at R (in both triangles) ≅ each other (vertical angles)
- Side QR ≅ SR — which is side opposite angle P in triangle PRQ, and side opposite angle T in triangle TRS.

So yes — two angles and a non-included side → AAS

Some curricula accept AAS as a valid congruence theorem.

So reason 5: AAS Congruence Theorem

✔ Second proof reasons filled.

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Third Proof:

Given:
- $\angle B \cong \angle E$
- $\angle BCA \cong \angle ECD$
- C is the midpoint of $\overline{AD}$
Prove: $\overline{AB} \cong \overline{DE}$

Step-by-step:

1. $\angle B \cong \angle E$ → Given
2. $\angle BCA \cong \angle ECD$ → Given
3. C is the midpoint of $\overline{AD}$ → Given
4. $\overline{AC} \cong \overline{DC}$ → Because C is midpoint → Definition of Midpoint
5. $\triangle ABC \cong \triangle DEC$ → We have:
- $\angle B \cong \angle E$ (angle)
- $\angle BCA \cong \angle ECD$ (angle)
- $\overline{AC} \cong \overline{DC}$ (side)

Now, is the side included? In triangle ABC, side AC is between angles B and BCA? No — side AC is between angles A and C.

Actually, we have two angles and a non-included side again? Let’s see:

In triangle ABC:
- Angles at B and C are known to be congruent to angles at E and C in triangle DEC.
- Side AC is in triangle ABC, side DC is in triangle DEC.

Side AC is adjacent to angle C and angle A — but we don’t know angle A.

Actually, side AC is between angles A and C — but we’re given angles B and C.

So side AC is NOT between the two given angles — so again, AAS.

Alternatively, since we have two angles, the third angle must also be congruent (because angles sum to 180), so we could use ASA — but that would require knowing the included side.

Actually, let’s check:

If we have:
- $\angle B \cong \angle E$
- $\angle BCA \cong \angle ECD$
- and side $\overline{AC} \cong \overline{DC}$

In triangle ABC, side AC is opposite angle B.

In triangle DEC, side DC is opposite angle E.

Since angle B ≅ angle E, and side AC ≅ DC, and another angle ≅, this fits AAS.

So → AAS Congruence Theorem

6. $\overline{AB} \cong \overline{DE}$ → Corresponding parts of congruent triangles → CPCTC

✔ Third proof reasons filled.

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Final Answers for each proof’s Reasons column:

First Proof Reasons:
1. Given
2. Given
3. Reflexive Property of Congruence
4. SAS Congruence Postulate
5. CPCTC

Second Proof Reasons:
1. Given
2. Given
3. Definition of Midpoint
4. Vertical Angles Theorem
5. AAS Congruence Theorem

Third Proof Reasons:
1. Given
2. Given
3. Given
4. Definition of Midpoint
5. AAS Congruence Theorem
6. CPCTC

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Final Answer:
[The completed reasons for all three proofs are as above.]