In Fig. 6.14, DE || QR and AP and BP are bisectors of ∠ EAB and ∠ RBA, respectively. Find ∠APB
Solution:
Given, DE is parallel to QR
AP and BP are the bisectors of ∠EAB and ∠RBA
We have to find the measure of ∠APB
We know that if two parallel lines are cut by a transversal, then the sum of interior angles lying on the same side of the transversal is always equal to 180 degrees.
∠EAB + ∠RBA = 180°
Dividing by 2 on both sides,
∠EAB/2 + ∠RBA/2 = 180°/2
∠EAB/2 + ∠RBA/2 = 90° -------------------- (1)
Since AP is the bisector of ∠EAB
∠EAP = ∠PAB
∠EAB = ∠EAP + ∠PAB
∠EAB = ∠PAB + ∠PAB
∠EAB = 2∠PAB
∠PAB = ∠EAB/2 ----------------------- (2)
Since BP is the bisector of ∠RBA
∠ABP = ∠PBR
∠RBA = ∠ABP + ∠PBR
∠RBA = ∠ABP + ∠ABP
∠RBA = 2∠ABP
∠ABP = ∠RBA/2 ------------- (3)
Adding (2) and (3),
∠PAB + ∠ABP = ∠EAB/2 + ∠RBA/2
From (1),
∠PAB + ∠ABP = 90° ------------- (4)
Considering triangle APB,
We know that the sum of all three interior angles of a triangle is always equal to 180 degrees.
∠PAB + ∠ABP + ∠APB = 180°
From (4),
90 + ∠APB = 180°
∠APB = 180°- 90°
Therefore, ∠APB = 90°
✦ Try This: Find the angles of 1, 2, 3, 5, 6 and 8
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 6
NCERT Exemplar Class 9 Maths Exercise 6.3 Problem 7
In Fig. 6.14, DE || QR and AP and BP are bisectors of ∠ EAB and ∠ RBA, respectively. Find ∠APB
Summary:
The figure represents two parallel lines cut by a transversal. In Fig. 6.14 DE || QR and AP and BP are bisectors of ∠ EAB and ∠ RBA, respectively. The measure of ∠APB is 90°
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