# In the given figure

Question:

In the given figure, $\angle 1=\angle 2$ and $\frac{A C}{B D}=\frac{C B}{C E}$.

Prove that ∆ ACB ∼ ∆ DCE.

Solution:

We have:

$\frac{A C}{R D}=\frac{C B}{C F}$

$\Rightarrow \frac{A C}{C B}=\frac{B D}{C E}$

$\Rightarrow \frac{A C}{C B}=\frac{C D}{C E}($ Since, $B D=D C$ as $\angle 1=\angle 2)$

Also, $\angle 1=\angle 2$

i.e, $\angle D B C=\angle A C B$

Therefore, by SAS similarity theorem, we get:

$\triangle A C B \sim \triangle D C E$

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