Question

The adjoining figure shows parallelogram ABCD. The bisectors of angles A, B, C and D
close a quadrilateral PQRS Using the exterior angle theorem, prove that each angle
of quadrilateral PQRS is a night angle​

Answer 1

The proof is given below:

∵ ABCD is a parallelogram

∴ AB ║ CD

And, AD ║ BC

We know that if a transverse line cuts two parallel lines, the sum of the interior angles on the same side of the line is 180°

∴ ∠D + ∠A = 180°        .......... (1)

But DQ is bisector of ∠D and AS is angle bisector of ∠A

∴ ∠PDA = ∠D/2

And, ∠PAD = ∠A/2

From (1)

(∠D + ∠A)/2 = 180°/2

or, ∠D/2 + ∠A/2 = 90°

or, ∠PDA + ∠PAD = 90°

In ΔADP

∠SPQ is exterior angle of Δ ADP

∴ ∠SPQ =  ∠PDA + ∠PAD    (∵ exterior angle is equal to the sum of internal angles)

∴ ∠SPQ = 90°

Similarly we can prove that

∠PSR = 90°

∠SRQ = 90°

∠RQP = 90°

Therefore, each angle of quadrilateral PQRS is a right angle.                      (Proved)

Know More:

Q: Prove that if the diagonals of a quadrilateral are equal and bisect each other at right angles,then it is a square.

Click here: https://brainly.in/question/887725

Q: The diagonals of a quadrilateral intersect at  right angles. Prove that the figure obtained by  joining the mid-points of the adjacent sides of  the quadrilateral is a rectangle.​

Click Here: https://brainly.in/question/12640173