If bac is 90degree AD is …

To prove the given result,we will use the following theorm.

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

Since AD is the bisector of {tex}\angle{/tex}A of {tex}\Delta{/tex}ABC.

{tex}\therefore \quad \frac { A B } { A C } = \frac { B D } { D C }{/tex} [by above theorm]
{tex}\Rightarrow \quad \frac { A B } { A C } + 1 = \frac { B D } { D C } + 1{/tex}[Adding 1 on both sides]
{tex}\Rightarrow \quad \frac { A B + A C } { A C } = \frac { B D + D C } { D C }{/tex}
{tex}\Rightarrow \quad \frac { A B + A C } { A C } = \frac { B C } { D C }{/tex} ... (i)
In {tex}\Delta{/tex}'s CDE and CBA, we have
{tex}\angle{/tex}DCE = {tex}\angle{/tex}BCA = {tex}\angle{/tex}C [Common]
{tex}\angle{/tex}BAC = {tex}\angle{/tex}DEC [Each equal to 90°]
So, by AA-criterion of similarity, we have
{tex}\Delta{/tex}CDE ~ {tex}\Delta{/tex}CBA
{tex}\Rightarrow \quad \frac { C D } { C B } = \frac { D E } { B A }{/tex}
{tex}\Rightarrow \quad \frac { A B } { D E } = \frac { B C } { D C }{/tex} ...(ii)
From (i) and (ii), we obtain
{tex}\frac { A B + A C } { A C } = \frac { A B } { D E } \Rightarrow D E \times ( A B + A C ) = A B \times A C{/tex}