Question

ABCD IS A SQUARE AND AC , BD THE TWO DIGONALS INTERSECT EACH OTHER AT O . IF P IS A POINT ON AB , SUCH THT AO= OP , PROVE THAT 3ANGLE POB = ANGLE AOP

Answer 1

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