In the given figure, $D$ is a point on side $BC$ of $∆ABC$, such that $\frac{BD}{CD}=\frac{AB}{AC}$. Prove that $AD$ is the bisector of $∆BAC$.
"
Given:
$D$ is a point on side $BC$ of $∆ABC$, such that $\frac{BD}{CD}=\frac{AB}{AC}$.
To do:
We have to prove that $AD$ is the bisector of $∆BAC$.
Solution:
Produce $BA$ to $E$ such that $AE=EC$ and join $EC$.
$\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}$
$\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AE}}$ ($\mathrm{AC}=\mathrm{AE}$)
This implies,
$DA \| CE$ (By converse of Thales Theorem)
Therefore,
$\angle 2 =\angle 3$........(i) (Alternate angles)
$\angle 1 =\angle 4$........(ii) (Corresponding angles)
$\mathrm{AE}=\mathrm{AC}$
This implies,
$\angle 3=\angle 4$.........(iii)
From (i), (ii) and (iii), we get,
$\angle 1=\angle 2$
This implies,
$\mathrm{AD}$ is the bisector of $\angle \mathrm{BAC}$.
Hence proved.
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