Question

In given figure the bisector of interior $\angle A$ of $△ABC$ meets BC in D, and the bisector of exterior $\angle A$ meets BC produced in E, prove that $\frac{BD}{BE}=\frac{CD}{CE}$.

Answer 1

Given, In $△ABC$ AD and AE are respectively the bisectors of the interior and exterior angles at A

To Prove $\frac{BD}{BE}=\frac{CD}{CE}$

Proof Since AD is the internal bisector of $\angle A$ meeting BC at D.

$\therefore$ $\frac{AB}{AC}=\frac{BD}{DC}$........(i)

Since AE is the external bisector of $\angle A$ meeting BC produced in E.

$\therefore$ $\frac{AB}{AC}=\frac{BE}{CE}$........(ii)

From (i) and (ii), we get

$\frac{BD}{DC}=\frac{BE}{CE}$

$\Rightarrow$ $\frac{BD}{BE}=\frac{CD}{CE}$ $\left[Henceproved\right]$