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

In square ABCD, AC is a diagonal.

So, ∠CAB = 45° (As diagonals bisect vertex angle)

∠OAP = 45°

In ∆AOP,

∠OAP = 45°

AO = AP [Given]

∠AOP = ∠APO = x (let) [Angles opposite to equal sides are equal]

Now,

∠AOP + ∠APO + ∠OAP = 180° [Angles sum property of a triangle]

∠AOP + ∠AOP + 45° = 180°

2x = 180° – 45°

x = $\dfrac{135°}{2}$

∠AOB = 90° [Diagonals of a square bisect at right angles]

So, ∠AOP + ∠POB = 90°

$\dfrac{135°}{2}$ + ∠POB = 90°

∠POB = 90° – $\dfrac{135°}{2}$

= $\dfrac{45°}{2}$

3∠POB = $3 \times \dfrac{45°}{2} = \dfrac{135°}{2}$ = ∠AOP.

Hence, proved that 3∠POB = ∠AOP.