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

Solution:

From the angle sum property of a quadrilateral, we know that the sum of the angles is 360º

∠A + ∠B + ∠C + ∠D = 360°

Dividing both LHS and RHS by 2

1/2 (∠A + ∠B + ∠C + ∠D) = 1/2 × 360° = 180°

AP, PB, RC and RD are the bisectors of ∠A, ∠B, ∠C and ∠D

∠PAB + ∠ABP + ∠RCD + ∠RDC = 180° …. (1)

We know that the sum of all angles of a triangle = 180°

∠PAB + ∠APB + ∠ABP = 180°

∠PAB + ∠ABP = 180° – ∠APB …. (2)

Similarly

∠RDC + ∠RCD + ∠CRD = 180°

∠RDC + ∠RCD = 180° – ∠CRD …. (3)

Let us substitute equation (2) and (3) in (1)

180° – ∠APB + 180° – ∠CRD = 180°

360° – ∠APB – ∠CRD = 180°

So we get

∠APB + ∠CRD = 360° – 180°

∠APB + ∠CRD = 180° …. (4)

∠SPQ = ∠APB and ∠SRQ = ∠DRC are vertically opposite angles

Substitute it in equation (4)

∠SPQ + ∠SRQ = 180°

So PQRS is a quadrilateral whose opposite angles are supplementary.

Therefore, PQRS is a quadrilateral whose opposite angles are supplementary.

✦ Try This: If bisectors of ∠P and ∠Q of a quadrilateral PQRS intersect each other at A, of ∠Q and ∠R at B, of ∠R and ∠S at C and of ∠S and ∠P at D, then ABCD is a a. rectangle, b. rhombus, c. parallelogram, d. quadrilateral whose opposite angles are supplementary

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 8

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NCERT Exemplar Class 9 Maths Exercise 8.1 Problem 7

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

Summary:

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 quadrilateral whose opposite angles are supplementary

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