Question

(B) In A ABC, AD and CE are medians intersecting each other at point G and DF | EC.
Prove that: EF = BF
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Answer 1

In ΔBFD and ΔBEC, ∠BFD = ∠BEC (Corresponding angles) ∠FBD = ∠EBC (Common) ΔBFD ~ ΔBEC (AA Similarity) (ii) In ΔAFD, EG || FD. Using Basic Proportionality theorem, Read more on Sarthaks.com - https://www.sarthaks.com/155558/in-the-following-figure-ad-and-ce-are-medians-of-abc-df-is-drawn-parallel-to-ce-prove-that

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