# Please prove statement

Here is the proof to your question.

The given information can be represented diagrammatically as:

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)]⇒ 2

*x*= 90 – 45 = 45⇒

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

⇒ ∠AOP = 3∠POB

Hope! You got the proof.

Cheers!

**
**