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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Rajan@2021Guru

Asked: March 25, 20212021-03-25T14:02:51+05:30 2021-03-25T14:02:51+05:30In: ICSE

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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MathsMentor · Answer 1 · 2021-03-28T15:02:36+05:30

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

In ΔBFD and ΔBEC

∠BFD=∠BEC (Corresponding angles)

∠FBD=∠EBC (Common angles)

ΔBFD∼ΔBEC (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,EG∣∣FD. 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