In Fig. 6.17, ∠Q > ∠R, PA is the bisector of ∠QPR and PM ⊥ QR. Prove that ∠APM = 1/2 (∠Q - ∠R)

Solution:

Given, ∠Q > ∠R

PA is the bisector of ∠QPR

PM ⊥ QR

We have to prove that ∠APM = 1/2 (∠Q - ∠R)

As PA is the bisector of ∠QPR

∠QPA = ∠APR ---------------------------- (1)

Considering triangle PQM,

By angle sum property,

∠PQM + ∠QPM + ∠PMQ = 180

Since PM is perpendicular to QR

∠M = 90

Now, ∠PQM + ∠QPM + 90 = 180

∠PQM + ∠QPM = 180 - 90

∠PQM + ∠QPM = 90

∠PQM = 90 - ∠QPM -------------------- (2)

Considering triangle PMR,

By angle sum property,

∠PMR + ∠PRM + ∠MPR = 180

90 + ∠PRM + ∠MPR = 180

∠PRM + ∠MPR = 180 - 90

∠PRM + ∠MPR = 90

∠PRM = 90 - ∠MPR ------------------- (3)

Subtracting (2) and (3),

∠PQM - ∠PRM = (90 - ∠QPM) - (90 - ∠MPR)

∠Q - ∠R = 90 - ∠QPM - 90 + ∠MPR

∠Q - ∠R = ∠MPR - ∠QPM

From the figure,

∠MPR = ∠MPA + ∠APR

Also, ∠QPA = ∠QPM + ∠APM

∠QPM = ∠QPA - ∠APM

Now, ∠Q - ∠R = ∠MPA + ∠APR - (∠QPA - ∠APM)

∠Q - ∠R = ∠MPA + ∠APR - ∠QPA + ∠APM

From (1),

∠Q - ∠R = ∠MPA + ∠APM + ∠QPA - ∠QPA

∠Q - ∠R = 2∠MPA

Therefore, ∠APM = 1/2 (∠Q - ∠R)

✦ Try This: In the given figure, I ∥ m, if s and t be transversals such that s is not parallel to t, find the values of x and y.

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 6

---

NCERT Exemplar Class 9 Maths Exercise 6.4 Problem 7

In Fig. 6.17, ∠Q > ∠R, PA is the bisector of ∠QPR and PM ⊥ QR. Prove that ∠APM = 1/2 (∠Q - ∠R)

Summary:

An angle bisector or the bisector of an angle is a ray that divides an angle into two equal parts. In Fig. 6.17, ∠Q > ∠R, PA is the bisector of ∠QPR and PM ⊥ QR. It is proven that ∠APM = 1/2 (∠Q - ∠R)

---

☛ Related Questions:

• Prove that through a given point, we can draw only one perpendicular to a given line
• Prove that two lines that are respectively perpendicular to two intersecting lines intersect each ot . . . .
• Prove that a triangle must have atleast two acute angles