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Question
Prove that the angle bisectors of the angles formed by producing opposite sides of a cyclic quadrilateral (Provided they are not parallel) intersect at the right angle.
Solution
Here, ABCD is a cyclic quadrilateral.
PM is the bisector of ∠ APB and QM is a bisector of ∠ AQD.
In Δ PDL and Δ PBN,
∠ 1 = ∠ 2 ...( PM is the bisector of ∠P )
∠ 3 = ∠ 9 ...( Exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.)
∠ 4 = ∠ 7
But, ∠ 4 = ∠ 8 ...( Vertically opposite angles)
∠ 7 = ∠ 8
Now in Δ QMN and Δ QML,
∠ 7 = ∠ 8 ...(prove above)
∠ 5 = ∠ 6 ...( QM is a bisector of Q)
Δ QMN ∼ Δ QML
∠ QMN = ∠ QML
But
∠ QMN + ∠ QML = 180°
∠ QMN = ∠ QML = 90°
Hence, ΔPMQ = 90° ...( ∵ ∠PMQ = ∠QML)
Hence proved.
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