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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.

Rectilinear Figures

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Answer

In square ABCD, AC is a diagonal.

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. Rectilinear Figures, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

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 = 135°2\dfrac{135°}{2}

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

So, ∠AOP + ∠POB = 90°

135°2\dfrac{135°}{2} + ∠POB = 90°

∠POB = 90° – 135°2\dfrac{135°}{2}

= 45°2\dfrac{45°}{2}

3∠POB = 3×45°2=135°23 \times \dfrac{45°}{2} = \dfrac{135°}{2} = ∠AOP.

Hence, proved that 3∠POB = ∠AOP.

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