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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)].

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

Answers (1)

Produce DE to intersect BC at P

Now, EF \parallel BC DP is the transversal
\angle DEF = \angle DPC   (corresponding angles) … (i)
Now AB \parallel DP BC is transversal,
\angle DPC = \angle ABC  (corresponding angles) … (ii)
From (i) and (ii) we get
\angle ABC = \angle DEF

Hence, Proved

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