Question

The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the corresponding sides containing the angle.Prove it.

Answer 1

Given,

Consider a triangle$ABC$
Let $AD$ be the internal bisector of $\angle BAC$ which meet$BC$ at$D$

To Prove

$BD/DC=AB/AC$

Construction

Draw $CE\parallel DA$ to meet $BA$produced at $E$

Proof

From the figure, we note that

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

So
$\angle DAC=\angle ACE$ (alternate interior angles angle ) ——(i)

$\angle BAD=\angle AEC$ (corresponding angle) ——(ii)

$AD$ is the angle bisector of $\angle A$

∴$\angle BAD=\angle DAC$——(iii)
From all the three equations above (i), (ii) and (iii) we conclude that

$\angle ACE=\angle AEC$

From$△ACE,$
$\Rightarrow AE=AC$[Sides opposite to equal angles are equal]

From $△BCE,$
$CE\parallel DA$

From Thales Theorem we know that the ratio of any two corresponding sides in two equiangular triangles is always the same. This theorem is called as Basic Proportionality Theorem

$$\begin{gathered} BD/DC=BA/AE \\ BD/DC=AB/AC \end{gathered}$$

Hence Proved.