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In Fig. 6.12, BA || ED and BC || EF. Show that ∠ABC = ∠DEF [Hint: Produce DE to intersect BC at P (say)]

Solution:

Given, the lines BA and ED are parallel.

Also, the lines BC and EF are parallel.

We have to show that ∠ABC = ∠DEF

Extend DE to meet BC at P.

In Fig. 6.12, BA || ED and BC || EF. Show that ∠ABC = ∠DEF [Hint: Produce DE to intersect BC at P

Now, EF || BC and DP is the transversal.

We know that if two parallel lines are cut by a transversal, then the corresponding angles are equal.

∠DEF = ∠CPD ---------------- (1)

Now, AB || DP and BC is the transversal.

So, ∠CPD = ∠ABC -------------- (2)

From (1) and (2),

∠DEF = ∠ABC

Therefore, it is proved that ∠DEF = ∠ABC

✦ Try This: In the given figure , the arms of two angles are parallel. If ∠ABC=70°, then find ∠DGC

In the given figure , the arms of two angles are parallel. If ∠ABC=70°, then find ∠DGC

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

NCERT Exemplar Class 9 Maths Exercise 6.3 Problem 5

Summary:

In Fig. 6.12, BA || ED and BC || EF. It is shown that ∠ABC = ∠DEF, as the parallel lines cut by transversal, corresponding angles are equal

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