In ∆ABC (Fig. 4.59), if ∠1 = ∠2, prove that ABAC=BDDC.


In $\triangle \mathrm{ABC}$ (Fig. 4.59), if $\angle 1=\angle 2$, prove that $\mathrm{ABAC}=\mathrm{BDDC}$.


We have to prove that $\frac{A B}{A C}=\frac{B D}{D C}$.

In ∆ABC,

$\angle \mathrm{l}=\angle 2$ (Given)

So, $A D$ is the bisector of $\angle A$

Therefore, $\frac{A B}{A C}=\frac{B D}{D C}$

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