Question

In the figure, AB = AC and CP || BA and AP is the bisector of exterior $\angle CADof\Delta ABC$. Prove that (i) $\angle PAC=\angle BCA$(ii) ABCP is a parallelogram.

Answer 1

Given : In ABC, AB = AC

and CP || BA, AP is the bisector of exterior

$\angle CADof\Delta ABC$

To prove :

(i) $\angle PAC=\angle BCA$

(ii) ABCP is a || gm

Proof : (i) In $\Delta ABC,$

$\because AB=AC$

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

and ext. $\angle CAD=\angle B+\angle C$

$=\angle C+\angle C=2\angle C$ ....(i)

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

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

From (i) and (ii)

$\angle C=2\angle PAC$

$\angle C=\angle CADor\angle BCA=\angle PAC$

Hence $\angle PAC=\angle BCA$

(ii) But there are alternate angles,

$\therefore$ AD|| BC

But BA||CP

$\therefore$ ABCP is a ||gm.