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Question
In the following figure, BA || ED and BC || EF. Show that ∠ABC = ∠DEF
[Hint: Produce DE to intersect BC at P (say)].
Solution
According to the question:
Given: Producing DE to intersect BC at P.
EF || BC and DP is the transversal,
∠DEF = ∠DPC ...(i) [Corresponding ∠s]
See the above figure, AB || DP and BC is the transversal,
∠DPC = ∠ABC ...(ii) [Corresponding ∠s]
Now, from equation (i) and (ii), we get:
∠ABC = ∠DEF
Hence proved.
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