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  • If bisectors of A and B of a quadrilateral ABCD intersect each other at P, of B and C at Q, of C and D at R and of D and A at S, then PQRS is a (A) rectangle (B) rhombus (C) parallelogram (D) quadrilateral whose opposite angles are supplementary

If bisectors of \angle A and \angle B of a quadrilateral ABCD intersect each other at P, of \angle B and \angle C at Q, of \angle C and \angle D at R and of \angle D and \angle A at S, then PQRS is a
(A) rectangle
(B) rhombus
(C) parallelogram

(D) quadrilateral whose opposite angles are supplementary

Answers (1)

First of all, let us draw the figure according to the question.

From the above diagram
\angle QPS=\angle APB           …..(1)             (Vertically opposite angles)
In \triangle APB=\angle APB+\angle PAB+\angle ABP=180^{\circ}
\angle APB=\frac{1}{2}\angle A+\frac{1}{2}\angle B=180^{\circ}
\angle APB=180-\frac{1}{2}(\angle A+\angle B)            …..(2)
From equation 1 & 2
\angle QPS=180-\frac{1}{2}(\angle A+\angle B)            …..(3)
Similarly  \angle QRS=180-\frac{1}{2}(\angle C+\angle D)       …..(4)
Add equations 3 and 4
\angle QPS+\angle QRS=360^{\circ}-\frac{1}{2}(\angle A+\angle B+\angle C+\angle D)
=360^{\circ}-\frac{1}{2}(360^{\circ})
=360^{\circ}-180^{\circ}
=180^{\circ}
\therefore \angle QPS+\angle QRS= 180^{\circ}
Hence PQRS is quadrilateral whose opposite angles are supplementary.

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