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Rajan@2021
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In the adjoining figure, medians AD and BE of △ABC meet at the point G, and DF is drawn parallel to BE. Prove that (i) EF=FC (ii) AG:GD=2:1.

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In the figure we have been asked to Prove that  (i) EF=FC (ii) AG:GD=2:1.

If Medians AD and BE intersect each other at G and a parallel line DF is drawn parallel to  BE

ML  Aggarwal Avichal Publication Similarity chapter 13 question no 19

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  1. In the following fig, AD and CE are median of ΔABC DF is drawn parallel to CE

    (1) EF=FB

    In ΔBFD and ΔBEC

    BFD=BEC (Corresponding angles)

    FBD=EBC (Common angles)

    ΔBFDΔBEC  (AA similarity)

    BF/BFE=BD/BC

    BF/BE=1/2(As is the mid point of BC)

    BE=2BF

    BF=FE=2BF

    Hence EF=FB

    ii) AG:GD=2:1
    In ΔAFD,EGFD. Using baise proportionality then:-

    AE​/EF=AG​/GD……..(1)

    Now AE=EB  (as E is the mid of AB)

    AE=2EF (Since EF=FB by 1)

    AG:GD=2:1……….From 1

    Hence AG:GD=2:1

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