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In the figure given below, $\angle 1 = \angle 2$ and $\frac{BE}{BC} = \frac{CD}{AB}$. Prove that $\triangle BDE \sim \triangle BAC$.
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Solution: In $\triangle DBC$, $\angle 1 = \angle 2$
$BD = CD$ (1/2 Mark)
In $\triangle BDE$ and $\triangle BAC$,
As $\frac{BE}{BC} = \frac{CD}{AB}$
$\therefore \frac{BE}{BC} = \frac{BD}{AB}$ (1/2 Mark)
$\angle B = \angle B$ (Common) (1/2 Mark)
$\therefore \triangle BDE \sim \triangle BAC$ (By SAS Similarity) (1/2 Mark)
$BD = CD$ (1/2 Mark)
In $\triangle BDE$ and $\triangle BAC$,
As $\frac{BE}{BC} = \frac{CD}{AB}$
$\therefore \frac{BE}{BC} = \frac{BD}{AB}$ (1/2 Mark)
$\angle B = \angle B$ (Common) (1/2 Mark)
$\therefore \triangle BDE \sim \triangle BAC$ (By SAS Similarity) (1/2 Mark)