Question

ABCD is a quadrilateral with AB=AD. The bisector of <BAC and <DAC intersect BC and CD at E and F respectively. Prove that EF// BD where AE=AF.

Answer 1

hey mate.
here's the proof

Answer 2

GIVEN:- AB=AD
To PROVE:- EF || BD

PROOF:-

In ∆ABC
AE is bisector Of <BAC

Using internal bisector theorem

$\frac{AC}{AB} = \frac{CE}{BE} - - - - - - - - 1 \\ similarly$

in ∆ ACD we have AF as bisector of <CAD


$\frac{AC}{AD} = \frac{CF}{FD} \\ or \\ \frac{AC}{AB} = \frac{CF}{FD} - - - - - 2$
Equating 1 and 2
we get
$\frac{CE}{BE} = \frac{CF}{FD}$
NOw in ∆ BCD we have
$\frac{CE}{BE} = \frac{CF}{BD}$
ACC to converse of BPT theorem

EF || BD