Question

In given figure ABCD is a quadrilateral; in which AB=AD. The bisector of $\angle BAC$ and $\angle CAD$ intersect the sides BC and CD at the points E and F respectively. Prove that EF||BD.

Answer 1

Given A quadrilateral ABCD in which AB=AD and the bisectors of $\angle BAC$ and $\angle CAD$ meet the sides BC and CD at E and F respectively.
To prove EF||BD

Construction Join AC, BD and EF.

Proof In $△CAB$, AE is the bisector of $\angle BAC$.

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

In $△ACD$, AF is the bisector of $\angle CAD$.

$\therefore$ $\frac{AC}{AD}=\frac{CF}{DF}$

$\Rightarrow$ $\frac{AC}{AB}=\frac{CF}{DF}$ [$\because$ AD=AB]........(ii)

From (i) and (ii), we get

$\frac{CE}{BE}=\frac{CF}{DF}$

$\Rightarrow$ $\frac{CE}{EB}=\frac{CF}{FD}$

Thus, in $△CBD$, E and F divide the sides CB and CD respectively in the same ratio. Therefore, by the converse of Thale's Theorem, we have
$EF||BD$