In Fig. 6.13, BA || ED and BC || EF. Show that ∠ ABC + ∠ DEF = 180°
Solution:
Given, the lines BA and ED are parallel
The lines BC and EF are parallel.
We have to show that ∠ ABC + ∠ DEF = 180°
Extend EF to meet AB at P.
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.
Since EF || BC and AB is transversal.
Sum of interior angles = ∠EPB + ∠PBC
∠EPB + ∠PBC = 180° ---------- (1)
Since AB || ED and PE is transversal
We know that if two parallel lines are cut by a transversal, then the corresponding angles are equal.
The corresponding angles are ∠EPB and ∠DEF
∠EPB = ∠DEF -------------------- (2)
Using (2) in (1),
∠DEF + ∠PBC = 180°
From the figure,
∠ABC = ∠PBC
Therefore, ∠ABC + ∠DEF = 180°
✦ Try This: P is a point on the bisector of ∠ABC. The line through P parallel to BA meets BC at Q. Prove that ΔBPQ is an isosceles triangle.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 6
NCERT Exemplar Class 9 Maths Exercise 6.3 Problem 6
In Fig. 6.13, BA || ED and BC || EF. Show that ∠ ABC + ∠ DEF = 180°
Summary:
Two or more lines that lie in the same plane and never intersect each other are known as parallel lines. In Fig. 6.13, BA || ED and BC || EF. It is shown that ∠ ABC + ∠ DEF = 180°
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