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)]
⇒ 2x = 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!

 

  • 0
What are you looking for?