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

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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Valentina 3 weeks 2021-09-04T17:30:48+00:00 2 Answers 0 views 0

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    0
    2021-09-04T17:32:41+00:00

    Answer:

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    0
    2021-09-04T17:32:44+00:00

    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]

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