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