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Question
Prove that the angles bisectors of the angles formed by producing opposite sides of a cyclic quadrilateral (provided they are not parallel) intersect at right triangle.
Solution
Here, ABCD is a cyclic quadrilateral. PM is bisector of ∠ APB and QM is bisector of ∠ AQD
In Δ PDL and Δ PBN, ∠ 1=∠ 2 (PM is a bisector of LP)
∠ 3 = ∠ 9 (exterior angle of cydic quad. = interior opposite angle)
∴ ∠ 4 = ∠ 7
But, ∠ 4 = ∠ 8 (vertical opposite angles)
∴ ∠ 7 = ∠ 8
Now in Δ QMN and Δ QML
∠ 7 = ∠ 8 (proved above)
∠ S = ∠ 6 (QM is bisector of Q)
∴ Δ QMN ~ Δ QML
⇒ ∠ QMN and ∠ QML
But, ∠ QMN + ∠ QML = 180°
∴ ∠ QMN = ∠ QML = 90°
Hence, ∠ PMQ = 90 °
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