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 1

$$\begin{gathered} \mathrm{In}△\mathrm{PAD}\mathrm{and}△\mathrm{PAB},\mathrm{we}\mathrm{have}: \\ \mathrm{AD}=\mathrm{AB}(\mathrm{Side}\mathrm{of}\mathrm{a}\mathrm{square}) \\ \mathrm{AP}=\mathrm{AP}\left(\mathrm{Common}\right) \\ \mathrm{PD}=\mathrm{PB}\left(\mathrm{Given}\right) \\ △\mathrm{PAD}\cong \mathrm{PAB}(\mathrm{SSS}\mathrm{criterion}) \\ \Rightarrow \angle \mathrm{APD}=\angle \mathrm{APB} \end{gathered}$$

$$\begin{gathered} \mathrm{In}△\mathrm{CPD}\mathrm{and}△\mathrm{CPB},\mathrm{we}\mathrm{have}: \\ \mathrm{CD}=\mathrm{CB}(\mathrm{Sides}\mathrm{of}\mathrm{square}) \\ \mathrm{CP}=\mathrm{CP}\left(\mathrm{Common}\right) \\ \mathrm{PD}=\mathrm{PB}\left(\mathrm{Given}\right) \\ △\mathrm{CPD}\cong △\mathrm{CPB}(\mathrm{SSS}\mathrm{test}) \\ \Rightarrow \angle \mathrm{CPD}=\angle \mathrm{CPB} \end{gathered}$$

$$\begin{gathered} \therefore \angle \mathrm{APD}+\angle \mathrm{CPD}=\angle \mathrm{APB}+\angle \mathrm{CPB} \\ \mathrm{But}\angle \mathrm{APD}+\angle \mathrm{CPD}+\angle \mathrm{APB}+\angle \mathrm{CPB}=360° \\ \Rightarrow \angle \mathrm{APD}+\angle \mathrm{CPD}=180° \end{gathered}$$

So, CPA is a straight line.
Hence, proved.