Congruent Triangles Proof Worksheet | PDF | Triangle Geometry ... - Free Printable

Let’s solve each problem step by step. We’ll fill in the blanks for Statements and Reasons, using what we know about triangle congruence.

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Problem 1:

Given:
- BC ≅ DC
- AC ≅ EC

Prove: ΔBCA ≅ ΔDCE

Looking at the diagram (even though we’re not describing it), points A, C, E are on one line; B, C, D on another — they cross at point C. So angle BCA and angle DCE are vertical angles.

Step-by-step:

1. First statement is given info → “BC ≅ DC; AC ≅ EC” → Reason: Given ✔

2. Second reason is “Vertical ∠s Theorem” → so the statement must be that the vertical angles are congruent → “∠BCA ≅ ∠DCE”

3. Now we have two sides and the included angle congruent → that’s SAS postulate!

So:

Statements:
1. BC ≅ DC; AC ≅ EC
2. ∠BCA ≅ ∠DCE
3. ΔBCA ≅ ΔDCE

Reasons:
1. Given
2. Vertical ∠s Theorem
3. SAS Postulate

✔ Done with Problem 1.

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Problem 2:

Given:
- JK ≅ LK
- JM ≅ LM

Prove: ΔKJM ≅ ΔKLM

We see both triangles share side KM. That’s a common side → reflexive property.

Step-by-step:

1. First statement should be the given info → “JK ≅ LK; JM ≅ LM” → Reason: Given

2. Second reason is “Reflexive Prop.” → so statement is “KM ≅ KM” (same segment)

3. Now we have three sides: JK≅LK, JM≅LM, KM≅KM → SSS postulate!

So:

Statements:
1. JK ≅ LK; JM ≅ LM
2. KM ≅ KM
3. ΔKJM ≅ ΔKLM

Reasons:
1. Given
2. Reflexive Prop.
3. SSS Postulate

✔ Done with Problem 2.

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Problem 3:

Given:
- ∠G ≅ ∠I
- FH bisects ∠GFI

Prove: ΔGFH ≅ ΔIFH

FH bisecting ∠GFI means it splits it into two equal angles → ∠GFH ∠IFH

Also, FH is shared by both triangles → reflexive property → FH ≅ FH

We already have ∠G ≅ ∠I (given), and now ∠GFH ≅ ∠IFH (from bisector), and side FH ≅ FH → that’s AAS (two angles and non-included side)

Wait — let’s check order:

In ΔGFH and ΔIFH:

- ∠G ≅ ∠I (given)
- ∠GFH ≅ ∠IFH (bisector)
- Side FH is common → but which side? It’s between the two angles in each triangle? Actually, no — in AAS, the side doesn’t need to be between them.

Actually, since we have two angles and a non-included side (FH is opposite ∠G and ∠I respectively), yes — AAS works.

But let’s fill the table as asked.

Statement 1: Given → “∠G ≅ ∠I; FH bisects ∠GFI” → Reason: Given

Statement 2: From bisector → “∠GFH ≅ ∠IFH” → Reason: Def. of Angle Bisector

Statement 3: Need reflexive prop → “FH ≅ FH” → Reason: Reflexive Prop.

Statement 4: Triangles congruent → “ΔGFH ≅ ΔIFH” → Reason: AAS Theorem

Note: Some curricula call this AAS, others might say ASA if you rearrange — but here, the side is not between the two angles, so AAS is correct.

Let’s confirm:

Angles: ∠G and ∠GFH in first triangle; ∠I and ∠IFH in second.

Side FH is adjacent to ∠GFH and ∠IFH, but opposite ∠G and I.

Since we have two angles and a non-included side → AAS.

Yes.

So:

Statements:
1. ∠G ≅ ∠I; FH bisects ∠GFI
2. ∠GFH ≅ ∠IFH
3. FH ≅ FH
4. ΔGFH ≅ ΔIFH

Reasons:
1. Given
2. Def. of Angle Bisector
3. Reflexive Prop.
4. AAS Theorem

✔ Done with Problem 3.

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Final Answer:

Problem 1:

Statements:
1. $\overline{BC} \cong \overline{DC}$; $\overline{AC} \cong \overline{EC}$
2. $\angle BCA \cong \angle DCE$
3. $\triangle BCA \cong \triangle DCE$

Reasons:
1. Given
2. Vertical Angles Theorem
3. SAS Postulate

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Problem 2:

Statements:
1. $\overline{JK} \cong \overline{LK}$; $\overline{JM} \cong \overline{LM}$
2. $\overline{KM} \cong \overline{KM}$
3. $\triangle KJM \cong \triangle KLM$

Reasons:
1. Given
2. Reflexive Property
3. SSS Postulate

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Problem 3:

Statements:
1. $\angle G \cong \angle I$; $\overline{FH}$ bisects $\angle GFI$
2. $\angle GFH \cong \angle IFH$
3. $\overline{FH} \cong \overline{FH}$
4. $\triangle GFH \cong \triangle IFH$

Reasons:
1. Given
2. Definition of Angle Bisector
3. Reflexive Property
4. AAS Theorem