Question

ABCD is a square and the diagonals intersect at O. If P is a point on AB such that AO=AP, prove that 3(angle POB)=(angle AOP).

Answer 1

ABCD is a square and AC, BD the 2 diagonals intersect each other at O. If P is a point on AB such that AO=OP, prove that 3of anglePOB=3of angleAOP.



Let ∠POB = x°
It is known that diagonals of a square bisect the angles.

∴ ∠OBP = ∠OAP = (90°/2) = 45°

Using exterior angle property for ∆OPB, ∠OPA = ∠OBP + ∠POB = 45° + x° In OAP, OA = AP .



∴ ∠OPA = ∠AOP ⇒ ∠AOP = 45° + x° …

(1) It is known that diagonals of square are perpendicular to each other.


∴ ∠AOP + ∠POB = 90° ⇒ 45° + x° + x° = 90°

[Using (1)]

⇒ 2x = 90 – 45 = 45 ⇒ x = 22.5

∴ ∠POB = 22.5° and ∠AOP = 45° + 22.5° = 67.5° = 3 × 22.5°

⇒ ∠AOP = 3∠POB