In a square ABCD, P is the midpoint of AD. BP and . . .

Question : In a square ABCD, P is the midpoint of AD. BP and CP are joined. Prove that ∠PCB=∠PBC. Also prove that if ∠PCB=45° then PC is perpendicular to PB.

Doubt by Parth 

Solution :

Given : ABCD is a square.
AP=DP

∠PCB=45°

To Prove : (i) ∠PCB=∠PBC
(ii) PC is perpendicular to PB.

Proof : 

(i) In ΔBAP and ΔCDP
AB=DC (opposite sides of square)
∠BAP=∠CDP [Each 90°]
AP=DP (Given)
ΔBAP≅ΔCDP (By SAS)
PB=PC (By CPCT)

In ΔBPC
PB=PC(Proved above)
∠PCB=∠PBC (Angles opposite to equal sides of a triangle are equal)

(ii) In ΔBPC

∠PCB+∠PBC+∠BPC=180°(ASP)
45°+45°+∠BPC=180°
90°+∠BPC=180°

∠BPC=180°-90°
∠BPC=90°

So, PC is perpendicular to PB.

Hence Proved.