Question

(those who will answer it i will mark as brainliest) In the figure, ∠CPD = ∠BPD and AD is the bisector of ∠BAC. Prove that i) ∆ CAP ≅ ∆ BAP and ii) CP = BP.

Answer 1

Step-by-step explanation:

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

To prove:

$$\begin{lgathered}\triangle \: CAP \cong \: \triangle \: BAP \\\end{lgathered}$$

Solution: Because AD is bisector of angle BAC

So,

$$\begin{lgathered}\angle \: PAC = \angle \: PAB \: \: \: \: ...eq1 \\\end{lgathered}$$

Now,

$$\begin{lgathered}\angle \: CPD \: = \angle \: BPD \: \: \: \: \: ...eq2 \\\end{lgathered}$$

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,

$$\begin{lgathered}\angle \: CPD = \angle \: PAC + \angle \: ACP \\\end{lgathered}$$

Same as in another ∆BAP

$$\begin{lgathered}\angle \: BPD = \angle \: PAB + \angle \: ABP \\ \\\end{lgathered}$$

As from eq1 and eq2,

we can say that

$$\begin{lgathered}\angle \: ABP = \angle \: ACP \: \: \: \: ...eq3 \\\end{lgathered}$$

Now, AP is common in both

So,

$$\begin{lgathered}\angle \: PAC = \angle \: PAB \\ \\ \angle \: ACP = \angle \: ABP \\ \\ AP = AP \: (common) \\ \\ by \: AAS \: criterion \: of \: congruency \: of \: triangle \\\end{lgathered}$$

$$\begin{lgathered}\triangle \: CAP \cong \: \triangle \: BAP \\ \\\end{lgathered}$$

Hope it helps you.

Answer 2

Answer:

hope it helps you friends