Question

In triangle ABC, D and E are points on the sides AB and AC respectively such that AD × EC = AE × DB. Prove that DE || BC.

Answer 1

Given AD×EC=AE×DB
⇒AD/DB=AE/EC
It meant that sides AB and AC are divided in the same ratio by DE.
We have,''If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side''.(Converse of Basic Proportionality theorem)
Therefore, we can say that DE║BC(third side).
Hence,proved.

Answer 2

★New Answer:-★

$\sf\:Given:-~ DE || BC \\\\</h3><p>\sf\: \frac{AD}{BD} = \frac{AE}{CE}........ (Theorem~ 6.1) \\\\ \sf\: { \therefore} \: \frac{BD}{AD} = \frac{CE}{AE} \\\\ \sf\: \frac{BD}{AD} + 1 = \frac{CE}{AE} + 1 \\\\ \sf\: \frac{BD + AD}{AD} = \frac{CE + AE}{AE} \\\\ \sf\: { \therefore} \: \red{\frac{AB}{AD} = \frac{AC}{AE}}...........(Proved!)$