In the given figure, ABCD is a square and P is a point inside it …

Question

In the given figure, $ABCD$ is a square and $P$ is a point inside it such that $PB = PD$. Prove that $CPA$ is a straight line.

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Answer

Given: $ABCD$ is a sqaure and $P$ is a point inside it such that $PB = PD$.

To Prove: $CPA$ is a straight line.
Proof: In $\triangle\text{APD}$ and $\triangle\text{APB}$
$\text{DA = AB}$ [$\therefore$ $ABCD$ is a square]
$\text{AP = AP}$ [Common]
and, $\text{PB = PD}$ [Given]
Thus by Side-Side-Side criterion of congruence, we have
$\triangle\text{APD}\cong\triangle\text{APB}$
The corresponding parts of the congruent triangles are equal.
$\therefore\angle\text{APD}=\angle\text{APB}...(1)$
Now consider the triangles, $\triangle\text{CPD}$ and $\triangle\text{CPB}.$
$\text{CD = CB}$ [$\therefore$ $ABCD$ is a square]
$\text{CP = CP}$ [Common]
and, $\text{PB = PD}$ [Given]
Thus by Side-Side-Side criterion of congruence, we have
$\triangle\text{CPD}\cong\triangle\text{CPB}$
The corresponding parts of the congruents triangle are equal.
Hence we have
$\angle\text{CPD}=\angle\text{CPB}....(2)$
Adding both sides of $(1)$ and $(2)$ we get
$\angle\text{APD}+\angle\text{CPD}=\angle\text{APB}+\angle\text{CPB}...(3)$
Angles aronud the point $P$ add upto $360^\circ$,
$\Rightarrow\angle\text{APD}+\angle\text{CPD}+\angle\text{APB}+\angle\text{CPB}=360^{\circ}$
$\Rightarrow\angle\text{APB}+\angle\text{CPB}=360^{\circ}-(\angle\text{APD}+\angle\text{CPD})...(4)$
Substituting $(4)$ in $(3)$ we get,
$\angle\text{APD}+\angle\text{CPD}=360^{\circ}-(\angle\text{APD}+\angle\text{CPD})$
i.e. $2(\angle\text{APD}+\angle\text{CPD})=360^{\circ}$
$\angle\text{APD}+\angle\text{CPD}=\frac{360}{2}=180^{\circ}$
This proves that $CPA$ is a straight line.