## In the figure , jf. seg AB || Seg DE, Seg AB congruent seg DE, Seg AC || Seg DF, seg AC congruent seg DF, then prove that se

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Step-by-step explanation:## 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]