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If a line intersects sides AB and AC of $\triangle$ ABC at D and E respectively, and is parallel to BC, prove that $\frac{AD}{AB} = \frac{AE}{AC}$.
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$\triangle ADE \sim \triangle ABC$
$\therefore \frac{AD}{AB} = \frac{AE}{AC}$_a_1.png)
$\therefore \frac{AD}{AB} = \frac{AE}{AC}$
_a_1.png)