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.

Answer 1

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 °