Question

In the given figure, AD is bisector of angle BAC and angle CPD = angle BPD. Prove that triangle CAP is congruent to triangle BAP


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Answer 1

Step-by-step explanation:

Given that: AD is bisector of angle BAC and angle CPD = angle BPD

To prove:

$\triangle \: CAP \cong \: \triangle \: BAP$

Solution: Because AD is bisector of angle BAC

So,

$\angle \: PAC = \angle \: PAB \: \: \: \: ...eq1$

Now,

$\angle \: CPD \: = \angle \: BPD \: \: \: \: \: ...eq2$

in ∆CAP ,angle CPD is external angle

From external angle property of triangle,external angle CPD is equal to sum of two opposite internal angles

So,

$\angle \: CPD = \angle \: PAC + \angle \: ACP$

Same as in another ∆BAP

$\angle \: BPD = \angle \: PAB + \angle \: ABP$

As from eq1 and eq2,

we can say that

$\angle \: ABP = \angle \: ACP \: \: \: \: ...eq3$

Now, AP is common in both

So,

$\angle \: PAC = \angle \: PAB \\ \\ \angle \: ACP = \angle \: ABP \\ \\ AP = AP \: (common) \\ \\ by \: AAS \: criterion \: of \: congruency \: of \: triangle$

$\triangle \: CAP \cong \: \triangle \: BAP$

Hope it helps you.

Answer 2

Right♥️

Step-by-step explanation:

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