GSE Geometry CPCTC Worksheet featuring triangle congruence problems with diagrams and proof exercises.

Let's solve each problem on the CPCTC Worksheet step by step, explaining the reasoning behind each answer.

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#1: ΔHEY ≅ ΔMAN by ________

We are given two triangles: ΔHEY and ΔMAN. We need to determine how they are congruent based on markings, then use CPCTC (Corresponding Parts of Congruent Triangles are Congruent) to find other congruent parts.

#### Step 1: Analyze the Diagram
- In ΔHEY:
- Side HE has a single tick mark.
- Angle at Y is marked with an arc.
- Angle at E is also marked with an arc.
- In ΔMAN:
- Side MA has a single tick mark.
- Angle at A is marked with an arc.
- Angle at N is marked with an arc.

Now match the markings:

- HE ≅ MA (both have one tick mark)
- ∠E ≅ ∠A (same arc marking)
- ∠Y ≅ ∠N (same arc marking)

So we have:
- Two angles and the included side?

Wait — let’s check the order.

Actually:
- ∠E ≅ ∠A
- HE ≅ MA
- ∠Y ≅ ∠N

But in triangle HEY, angle at E and angle at Y are adjacent to side HE.

In triangle MAN, angle at A and angle at N are adjacent to side MA.

So:
- ∠E ≅ ∠A
- HE ≅ MA
- ∠Y ≅ ∠N

This matches ASA (Angle-Side-Angle) because:
- Angle at E → Angle at A
- Side HE → Side MA
- Angle at Y → Angle at N

So, ΔHEY ≅ ΔMAN by ASA.

> ✔ Answer: ASA

Now, using CPCTC, the remaining corresponding parts are congruent:

- HY ≅ MN (remaining sides)
- EY ≅ AN (other sides)
- ∠H ≅ ∠M (remaining angles)

So fill in:

```
HY ≅ MN
EY ≅ AN
∠H ≅ ∠M
```

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#2: ΔCAT ≅ ________, by ________

We are given a diagram with two triangles: CAT and RAP? Let's look closely.

Points:
- Triangle CAT: C, A, T
- Triangle RAP: R, A, P

They share point A, and lines cross at A.

Markings:
- ∠C ≅ ∠R (arc marks)
- ∠T ≅ ∠P (arc marks)
- CA ≅ RA (one tick mark)
- TA ≅ PA (one tick mark)

So:
- ∠C ≅ ∠R
- CA ≅ RA
- ∠T ≅ ∠P

Wait — that's two angles and a side.

But the side is not between the angles? Let's see:

In ΔCAT:
- ∠C and ∠T are at vertices C and T
- Side CT is opposite A
- But we have CA and TA marked as congruent

CA and TA are sides from A to C and A to T.

Similarly, RA and PA are from A to R and A to P.

So:
- CA ≅ RA
- TA ≅ PA
- ∠C ≅ ∠R
- ∠T ≅ ∠P

Also, since vertical angles at A are equal (∠CAT and ∠RAP), but wait — the diagram shows two intersecting lines: CT and RP cross at A.

So ∠CAT and ∠RAP are vertical angles, so they are congruent.

But looking at the arcs:
- ∠C ≅ ∠R
- ∠T ≅ ∠P
- And the side between them?

Wait — actually, let’s consider triangle CAT and triangle RAP?

But the question says: ΔCAT ≅ ________, by ________

Looking at the figure:
- Point A is common
- Lines CT and RP cross at A
- Markings:
- CA ≅ RA (tick marks)
- TA ≅ PA (tick marks)
- ∠C ≅ ∠R
- ∠T ≅ ∠P

So in ΔCAT and ΔRAP:
- CA ≅ RA
- TA ≅ PA
- ∠C ≅ ∠R
- ∠T ≅ ∠P

So we have two sides and the included angle? Wait — no.

Actually, the angles at C and T are not adjacent to the same side.

But we can use SAS or ASA?

Wait — better to use ASA:

Look:
- ∠C ≅ ∠R
- CA ≅ RA
- ∠A (in ΔCAT) and ∠A (in ΔRAP) — are they congruent?

But ∠CAT and ∠RAP are vertical angles → they are congruent.

So:
- ∠C ≅ ∠R (given)
- CA ≅ RA (given)
- ∠A ≅ ∠A (vertical angles)

So by ASA:
- ΔCAT ≅ ΔRAP

> ✔ Answer: ΔCAT ≅ ΔRAP by ASA

Now, by CPCTC, the remaining parts are congruent:

- CT ≅ RP (third side)
- ∠T ≅ ∠P (already known, but now confirmed)
- ∠A ≅ ∠A (but this is already used)

Wait — we need to list three pairs:

From CPCTC:
- CT ≅ RP
- AT ≅ AP (but already given)
- AC ≅ AR (already given)

But the blanks are:

```
______ ≅ ______, by CPCTC
______ ≅ ______, by CPCTC
______ ≅ ______, by CPCTC
```

So we should write the corresponding parts that are not already given.

Since ΔCAT ≅ ΔRAP, the correspondence is:
- C ↔ R
- A ↔ A
- T ↔ P

So:
- CT ≅ RP
- ∠T ≅ ∠P
- ∠C ≅ ∠R

But ∠C ≅ ∠R and ∠T ≅ ∠P were given, so maybe we just state all three.

But usually, CPCTC is used to conclude new congruences.

So likely:
- CT ≅ RP (new)
- ∠T ≅ ∠P (already marked, but can be stated)
- ∠C ≅ ∠R (same)

But perhaps the expected answers are:

```
CT ≅ RP, by CPCTC
∠T ≅ ∠P, by CPCTC
∠C ≅ ∠R, by CPCTC
```

Even though some were given, CPCTC confirms them.

Alternatively, if only non-given parts are required, then just CT ≅ RP.

But the worksheet has three blanks, so probably all three.

> ✔ Final Answer for #2:
> - ΔCAT ≅ ΔRAP, by ASA
> - Therefore:
> - CT ≅ RP, by CPCTC
> - ∠T ≅ ∠P, by CPCTC
> - ∠C ≅ ∠R, by CPCTC

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#3: Prove ∠3 ≅ ∠4

Given:
- AC ≅ AR
- ∠1 ≅ ∠2

Prove:
- ∠3 ≅ ∠4

Diagram: Two triangles, ΔLCA and ΔSRA, crossing at point A.

Lines LC and SR cross at A, forming two triangles: LCA and SRA.

We are to prove ∠3 ≅ ∠4.

#### Proof Steps:

We are given:
1. AC ≅ AR → Given
2. ∠1 ≅ ∠2 → Given

We are told:
3. ∠CAL ≅ ∠RAS → Why?

Because these are vertical angles — formed by intersecting lines LC and SR at point A.

So:
- ∠CAL and ∠RAS are vertical angles → they are congruent.

So Step 3: ∠CAL ≅ ∠RAS → Vertical Angles Theorem

Then:
4. ΔLCA ≅ ΔSRA → By ASA?

Let’s see:

In ΔLCA and ΔSRA:
- ∠1 ≅ ∠2 (given)
- AC ≅ AR (given)
- ∠CAL ≅ ∠RAS (vertical angles)

So:
- ∠1 and ∠CAL are adjacent to side AC
- ∠2 and ∠RAS are adjacent to side AR

So we have:
- ∠LAC = ∠1 + ∠CAL
- ∠SAR = ∠2 + ∠RAS

But wait — actually, the angles are:
- In ΔLCA: ∠L (which is ∠1), side AC, and ∠CAL
- In ΔSRA: ∠S (which is ∠2), side AR, and ∠RAS

So:
- ∠1 ≅ ∠2 (given)
- AC ≅ AR (given)
- ∠CAL ≅ ∠RAS (vertical angles)

So by ASA: two angles and the included side.

Yes! So ΔLCA ≅ ΔSRA by ASA

Then:
5. ∠3 ≅ ∠4 → by CPCTC, since they are corresponding angles.

Now fill in the proof table:

| Statement | Reason |
|---------|--------|
| 1. AC ≅ AR | Given |
| 2. ∠1 ≅ ∠2 | Given |
| 3. ∠CAL ≅ ∠RAS | Vertical Angles Theorem |
| 4. ΔLCA ≅ ΔSRA | ASA |
| 5. ∠3 ≅ ∠4 | CPCTC |

> ✔ Final Answers:

Proof:
1. $\overline{AC} \cong \overline{AR}$ → Given
2. $\angle 1 \cong \angle 2$ → Given
3. $\angle CAL \cong \angle RAS$ → Vertical Angles Theorem
4. $\triangle LCA \cong \triangle SRA$ → ASA
5. $\angle 3 \cong \angle 4$ → CPCTC

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✔ Final Answers Summary:

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#1

- ΔHEY ≅ ΔMAN by ASA
- Other congruent parts by CPCTC:
- $ \overline{HY} \cong \overline{MN} $
- $ \overline{EY} \cong \overline{AN} $
- $ \angle H \cong \angle M $

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#2

- ΔCAT ≅ ΔRAP, by ASA
- Therefore:
- $ \overline{CT} \cong \overline{RP} $, by CPCTC
- $ \angle T \cong \angle P $, by CPCTC
- $ \angle C \cong \angle R $, by CPCTC

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#3: Proof

1. $ \overline{AC} \cong \overline{AR} $ → Given
2. $ \angle 1 \cong \angle 2 $ → Given
3. $ \angle CAL \cong \angle RAS $ → Vertical Angles Theorem
4. $ \triangle LCA \cong \triangle SRA $ → ASA
5. $ \angle 3 \cong \angle 4 $ → CPCTC

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