If bisectors of `angleA and angleB ` of a quadrilateral ABCD intersect each other at P, of `angleB and angleC` at Q, of `angleC and angleD` at R and of `angleD and angleA` at S, then PQRS is a

Options

A. rectangle
B. rhombus
C. parallelogram
D. quadrilateral whose opposite angles are supplementary

Answer

D

Explanation

Given, ABCD is a quadrilateral and all angles bisectors form a quadrilateral PQRS.
ltBrgt We know that, sum of all angles in a quadrilateral is `360^(@)`.
`therefore" "angleA+angleB+angleC+angleD=360^(@)`
On dividing both sides by 2, we get
`(1)/(2)(angleA+angleB+angleC+angleD)=(360^(@))/(2)` ltBrgt `rArr" "anglePAB+anglePBA+angleRCD+angleRDC=180^(@)" "...(i)`
[since, AP and PB are the bisectors of `angleA and angleB` respectively also RC and RD are the bisectors of `angleC and angleD` respectively]
Now, in `Delta`APB, ltBrgt `" "anglePAB+angleABP+angleBPA=180^(@)`
`" "` [by angle sum properly of a triangle]
`rArr" "anglePAB+angleABP=180^(@)-angleBPA" "...(ii)`
Similarly in `Delta`RDC, ltBrgt `angleRDC+angleDCR+angleCRD=180^(@)` ltBrgt `" "` [by angle sum property of a triangle]
`rArr" "angleRDC+angleDCR=180^(@)-angleCRD" "...(iii)`
On substituting the value Eqs. (ii) and (iii) in Eq. (i), we get ltBrgt `180^(@)-angleBPA+180^(@)-angleDRC=180^(@)`
`rArr" "angleBPA+angleDRC=180^(@)` ltBrgt `rArr" " angleSPQ+angleSRQ=180^(@)`
`" "[becauseangleBPA=angleSPQandangleDRC=angleSRQ` vertically opposite angles]
Hence, PQRS is a quadrilateral whose opposite angles are suppolementary.