Question

State whether following statement is true or false.

The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle.

• A. True
• B. False

Answer 1

The correct option is A True

In $△ABC,AD$ is the external bisector of $\angle BAC$ and intersects $BC$ produced at $D$

Draw $CE\parallel DA$ meeting $AB$ at $E$

$CE\parallel DA$ and $AC$ is the transversal.

$\angle ECA=\angle CAD$ ....$\left(1\right)\because$ alternating angles

Also $CE\parallel DA$ and $BP$ is a transversal.

$\angle CEA=\angle DAP$.....$\left(2\right)\because$ corresponding angles.

But $AD$ is the bisector of $\angle CAP$

$\angle CAD=\angle DAP$ .........$\left(3\right)$

From $\left(1\right),\left(2\right)$ and $\left(3\right)$ we have

$\angle CEA=\angle ECA$(sides opposite to equal angles are equal)

In $△BDA,$ we have $EC\parallel AD$

$\therefore \frac{BD}{DC}=\frac{BA}{AE}$(by thales theorem)

$\therefore \frac{BD}{DC}=\frac{BA}{AC}$ since $AE=AC$

Hence proved.