Mathematics
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.
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 =
∠AOB = 90° [Diagonals of a square bisect at right angles]
So, ∠AOP + ∠POB = 90°
+ ∠POB = 90°
∠POB = 90° –
=
3∠POB = = ∠AOP.
Hence, proved that 3∠POB = ∠AOP.
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