Question

In the given figure, AD is bisector of ∠BAC and ∠CPD = ∠BPD. Prove that ΔCAP ≌ ΔBAP.

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Answer 1

Given :

• ∠CPD = ∠BPD

To prove :

• ΔCAP ≌ ΔBAP

Proof :

In ΔCAP & ΔBAP,

• AP = AP ( common side )
• ∠CPA = ∠BPA ( Since ∠CPD = ∠BPD so, ∠CPD + ∠CPA = ∠BPD + ∠BPA =>∠CPD + ∠CPA = ∠CPD + ∠BPA => ∠CPA = ∠BPA)
• ∠BAP = ∠CAP ( AD bisected ∠BAC leaving it with two equal angles )

Therefore,

ΔCAP ≌ ΔBAP by Angle-Side-Angle postulate.

Hence, proved ✓

Answer 2

Therefore, ΔCAP ≌ ΔBAP is proven.

Explanation:

To prove that ΔCAP ≌ ΔBAP, we need to show that they are congruent triangles.

Given:

1. AD is the bisector of ∠BAC.

2. ∠CPD = ∠BPD.

To prove:

ΔCAP ≌ ΔBAP.

Proof:

1. Given that AD is the bisector of ∠BAC, we know that ∠CAD = ∠BAD (by definition of an angle bisector).

2. Since ∠CPD = ∠BPD (given), we can conclude that ∠CPD + ∠CPA = ∠BPD + ∠BPA.

3. Combining the above equation with ∠CPA + ∠APD = ∠BPA + ∠BPD, we get ∠CPA + ∠APD = ∠APD + ∠BPA.

4. By rearranging the terms, we have ∠CPA = ∠BPA.

5. From step 4, we can conclude that ∠CAP = ∠BAP (since ∠CAD = ∠BAD).

6. We also know that AC = AB (common side).

7. Using the Angle-Side-Angle (ASA) congruence criterion, ΔCAP ≌ ΔBAP (by matching angles and the common side).

Therefore, ΔCAP ≌ ΔBAP is proven.