Question

In a square ABCD, Diagonals meet at O. P is a point on BC,such that OB=BP.Show that: angle poc =22.5

Answer 1

Given:-ABCD is a square.
OB=OP
To prove:-∠POC=22.5°
proof:-
in ΔAOB & ΔBOC
AO=OC[∵diagonals of a square perpendicularly bisect each other]
AB=BC[∵sides of a square]
OB is common
∴ΔAOB≅ΔBOC[SSS]
∠ABO=∠CBO[∵CPCT].......(1)
also
∠ABO+∠CBO=∠ABC
⇒∠ABO+∠CBO=90°[∵ angles of a square]
⇒2∠OBC=90°[using(1)]
∠OBC=45°.......(2)
in ΔOBC
OB=OP[given]
∠BOP=∠BPO....(3)
also
∠OBC+∠BOP+∠BPO=180° ⇒2∠BOP=180°-45[using(2)&(3)]
⇒∠BOP=135°/2
⇒∠BOP=67.5°........(4)
now
∠BOP+∠POC=∠BOC
⇒∠POC=90°-67.5[∵∠BOC=90°{∵diagonals square perpendicularly bisect each other} and by using (4)]
⇒∠POC=22.5°
∴∠POC=22.5°[proved]