Question

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 seg BC || seg Ef and seg BC congruent seg EF.
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Answer 1

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]

Answer 2

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