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Question

In the given figure, AB || CD. Prove that ∠BAE − ∠DCE = ∠AEC.

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Solution


Draw EFABCD through E.
Now, EFAB and AE is the transversal.
Then, BAE+AEF=180° Angles on the same side of a transversal line are supplementary
Again, EFCD and CE is the transversal.
Then,
DCE+CEF=180° Angles on the same side of a transversal line are supplementaryDCE+AEC+AEF=180°DCE+AEC+180°-BAE=180°BAE-DCE=AEC

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