Question

in the given figure PQRS is a ||gm with bisectors PA,QD,RC,SB respectively of angles P ,Q,R,S .Show that ABCD is a rectangle

Answer 1

in this ||gm
angle p+ q =180
then half of it will 90
thus applying angle property of triangle we find angle d =90
similarly all angle a,b,c, be 90
thus abcd be a rectangle

Answer 2

Proved below.

Step-by-step explanation:

Given:

PQRSis a parallelogram. PA bisects ∠SPQ. QD bisects ∠PQR. RC bisects ∠SRQ and SB bisects ∠PSR

To prove: ABCB is a rectangle

∠SPQ + ∠PQR = 180° because adjacent angles of a parallelogram are supplementary

∠QPA=$\frac{1}{2}$ SPQ   [PA bisects ∠SPQ ]

∠PQD=$\frac{1}{2}$ PQR   [QD bisects ∠PQR]

∠QPA+∠PQD= $\frac{1}{2}$ SPQ+$\frac{1}{2}$ PQR

∠QPA+∠PQD= $\frac{1}{2}$ (∠SPQ+∠PQR)

∠QPA+∠PQD= $\frac{1}{2}$ (180°)   [Since sum of adjacent angles of a parallelogram are supplementary]

ΔPQD is a right triangle since its acute interior angles are complementary

Similarly in ΔRSB we get ∠RSB = 90° and in ΔPAS we get ∠PAS = 90°

Then

∠BAD = 90° as ∠PAS and ∠DAB are vertical opposite angles

Since three angles of quadrilateral ABCD are right angles, hence 4th angle is also a right angle.

Thus ABCD is a rectangle.