AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (Fig. 6.11). Show that AP || BQ
Solution:
Given, two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m.
AP and BQ are the bisectors of the two alternate interior angles.
We have to show that AP || BQ
Given, the two parallel lines l and m are cut by a transversal t.
We know that if a transversal intersects two parallel lines, then alternate interior angles are equal.
From the figure,
The alternate interior angles are ∠EAB and ∠ABH
So, ∠EAB = ∠ABH -------- (1)
Dividing (1) by 1/2 on both sides,
∠EAB/2 = ∠ABH/2 ----------------- (2)
Since AP is the bisector of ∠EAB
∠EAP = ∠PAB
∠EAB = ∠EAP + ∠PAB
∠EAB = ∠PAB + ∠PAB
∠EAB = 2∠PAB
∠PAB = ∠EAB/2 -------------- (3)
Since BQ is the bisector of ∠ABH
∠ABQ = ∠HBQ
∠ABH = ∠ABQ + ∠HBQ
∠ABH = ∠ABQ + ∠ABQ
∠ABH = 2∠ABQ
∠ABQ = ∠ABH/2 -------------- (4)
Using (3) and (4) in (2) we get,
∠PAB = ∠ABQ
∠PAB and ∠ABQ are the alternate interior angles of the lines AP and BQ with transversal AB
Therefore, the lines AB and BQ are parallel.
✦ Try This: In the given figure, if line AB ∣∣ line CF and line BC ∣∣ line ED then prove that ∠ABC=∠FDE.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 6
NCERT Exemplar Class 9 Maths Exercise 6.3 Problem 3
AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (Fig. 6.11). Show that AP || BQ
Summary:
AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (Fig. 6.11). It is shown that AP || BQ as AP and BQ are two lines cut by a transversal AB
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