Question

In Figure, $ \mathrm{D}$ is a point on side $ \mathrm{BC}$ of $ ∆\mathrm{ABC}$ such that $ \frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{AB}}{\mathrm{AC}}.$ Prove that $ \mathrm{AD}$ is the bisector of$ \angle \mathrm{BAC}$.

Answer 1

Prove that $\mathrm{AD}$ is the bisector of $\angle \mathrm{BAC}.$

Given: $\frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{AB}}{\mathrm{AC}}$

Construction: Extends line $\mathrm{AB}$ such that $\mathrm{AP}=\mathrm{AC}$ and join $\mathrm{PC}$

$\mathrm{In}∆\mathrm{APC}$

$\mathrm{AP}=\mathrm{AC}$

$\therefore \angle \mathrm{APC}=\angle \mathrm{ACP}$ (Angle opposite to equal sides are equal)

Now,

$\frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{BA}}{\mathrm{AC}}$ (Given)

$\frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{BA}}{\mathrm{AP}}(\because \mathrm{AC}=\mathrm{AP})$

Now, in triangle $\mathrm{BPC},\mathrm{AD}$ is the line that divides the two sides $\mathrm{BC}$ and $\mathrm{BP}$ in the same ratio.

From the converse of the Basic Proportional Theorem, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

$\therefore \mathrm{DA}\parallel \mathrm{CP}$

$\angle \mathrm{BAD}=\angle \mathrm{APC}$ (corresponding angles)

$\angle \mathrm{DAC}=\angle \mathrm{ACP}-\left(1\right)$ (alternate angles)

And,

$$\begin{gathered} \angle \mathrm{BAD}=\angle \mathrm{ACP}(\angle \mathrm{APC}=\angle \mathrm{ACP}) \\ \angle \mathrm{BAD}=\angle \mathrm{DAC}\left(From1\right) \end{gathered}$$

Hence $\mathrm{AD}$ is the bisector of angle $\mathrm{BAC}$

Hence proved $\mathrm{AD}$ is the bisector of angle $\mathrm{BAC}$.