Question

In Fig., AB = AC and $CP\parallel BA$ and AP is the bisector of exterior $\angle CAD$ of $\Delta ABC$. Prove that $ABCP$ is a parallelogram.

Answer 1

We have ,

$\Rightarrow$ $AB=AC$ [ Given ]

$\Rightarrow$ $\angle C=\angle B$ ----- ( 1 ) [ Given ]

Now,

$\Rightarrow$ $\angle CAD=\angle B+\angle C$ [ Exterior angle theorem ]

$\Rightarrow$ $\angle CAD=\angle C+\angle C$ [ From eq ( 1 ) ]

$\Rightarrow$ $2\angle PAC=2\angle C$ [ $AP$ is the bisector of exterior $\angle CAD$ ]

$\Rightarrow$ $\angle PAC=\angle C$ [Alternate angles are equal ]

$\Rightarrow$ $\angle PAC=\angle BCA$

$\therefore$ $AP\parallel BC$

But, we have $AB\parallel CP$

As we know that if both pairs of opposite sides are parallel of a quadrilateral then it is a parallelogram.

Hence $ABCP$ is a parallelogram.