In the figure, $AB = AC$ and $CP \parallel BA$ and $AP$ is the bisector of exterior $\angle CAD$ of $\triangle ABC$. Prove that $\angle PAC = \angle BCA$.
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Given:

In the figure, $AB = AC$ and $CP \parallel BA$ and $AP$ is the bisector of exterior $\angle CAD$ of $\triangle ABC$.

To do:

We have to prove that $\angle PAC = \angle BCA$.

Solution:

In $\triangle ABC$,

$AB =AC$

This implies,

$\angle C = \angle B$                   (Angles opposite to equal sides)

$\angle CAD = \angle B + \angle C$

$= \angle C + \angle C$

$= 2\angle C$.......….(i)

$AP$ is the bisector of $\angle CAD$

This implies,

$2\angle PAC = \angle CAD$.........…(ii)

From (i) and (ii), we get,

$\angle C = 2\angle PAC$

$\angle C = \angle CAD$

$\angle BCA = \angle PAC$

Hence, $\angle PAC = \angle BCA$.

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Updated on: 10-Oct-2022

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