$ABC$ is an isosceles triangle in which $AB = AC$. $AD$ bisects exterior angle $PAC$ and $CD \parallel AB$ (see figure). Show that $\angle DAC = \angle BCA$.
(N/A) In $\Delta ABC$,we have $AB = AC$ (Given).
So,$\angle ABC = \angle ACB$ (Angles opposite to equal sides).
Also,$\angle PAC = \angle ABC + \angle ACB$ (Exterior angle property of a triangle).
Since $\angle ABC = \angle ACB$,we can write:
$\angle PAC = \angle ACB + \angle ACB = 2 \angle ACB$ ........... $(1)$
Now,$AD$ bisects $\angle PAC$,which means:
$\angle PAC = 2 \angle DAC$ ........... $(2)$
From equations $(1)$ and $(2)$,we get:
$2 \angle DAC = 2 \angle ACB$
Therefore,$\angle DAC = \angle ACB$ or $\angle DAC = \angle BCA$.
So,$\angle ABC = \angle ACB$ (Angles opposite to equal sides).
Also,$\angle PAC = \angle ABC + \angle ACB$ (Exterior angle property of a triangle).
Since $\angle ABC = \angle ACB$,we can write:
$\angle PAC = \angle ACB + \angle ACB = 2 \angle ACB$ ........... $(1)$
Now,$AD$ bisects $\angle PAC$,which means:
$\angle PAC = 2 \angle DAC$ ........... $(2)$
From equations $(1)$ and $(2)$,we get:
$2 \angle DAC = 2 \angle ACB$
Therefore,$\angle DAC = \angle ACB$ or $\angle DAC = \angle BCA$.
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