If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form
a. a square
b. a rhombus
c. a rectangle
d. any other parallelogram

Solution:

Consider the bisectors of angles APQ and CPQ meet at the point M and the bisectors of angles BPQ and PQD meet at the point N

Now join PM, MQ, QN and NP

As APB || CQD

∠APQ = ∠PQD

As NP and PQ are angle bisectors

2∠MPQ = 2 ∠NQP

Let us divide both sides by 2

∠MPQ = ∠NQP

PM || QN

In the same way,

∠BPQ = ∠CQP

PN || QM

So PNQM is a parallelogram

We know that angles on a straight line is 180°

∠CQP + ∠CQP = 180°

2∠MPQ + 2∠NQP = 180°

By dividing both sides by 2

∠MPQ + ∠NQP = 90°

∠MQN = 90°

So PMQN is a rectangle.

Therefore, the bisectors of the angles APQ, BPQ, CQP and PQD form a rectangle.

✦ Try This: If DPE and FQH are two parallel lines, then the bisectors of the angles DPQ, EPQ, FQP and PQH form a. a square, b. a rhombus, c. a rectangle, d. any other parallelogram

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

---

NCERT Exemplar Class 9 Maths Exercise 8.1 Problem 8

If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form , a. a square, b. a rhombus, c. a rectangle, d. any other parallelogram

Summary:

If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form a rectangle

---

☛ Related Questions:

• The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is ,a. a rh . . . .
• D and E are the mid-points of the sides AB and AC of ∆ABC and O is any point on side BC. O is joined . . . .
• The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is . . . .