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