ABCD is a square and AC, BD the two diagonals intersect each other at O. If P is a point on AB, such that AO = AP, prove that 3∠POB = ∠AOP.
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The given information can be represented diagrammatically as:
![](https://s3mn.mnimgs.com/img/shared/userimages/mn_images/image/02Feb2011-3.png)
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
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