Question

In quadrilateral ABCD , side AB congruent to side AD . Bisector of angle BAC cuts side BC at E and Bisector of angle DAC cuts side DC at F. prove that seg EF || seg BD

Answer 1

In ΔABC AE is the bisector of ∠BAC.

AC/AB = CE/BE -- 1 ( Using internal bisector theorum that states the angle bisector divides the opposite sides in ratio of sides consisting of the angles.)

In ΔACD AF is the bisector of ∠CAD

AC/AD = CF/FD

AC/AB = CF/FD -- 2

Equating 1 and 2 -  

CE/BE = CF/FD

In ΔBCD we have -  

CE/BE = CF/FD

Therefore, according to the converse of BPT theorem, we will get -  

If a line divides any two sides of a triangle in the same ratio, then the line should be parallel to its third side.

= EF || BD  [Hence Proved]

Answer 2

Answer:

Step-by-step explanation: