In quadrilateral abcd,side ab congruent side ad.Bisector of angle bac
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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]
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