Question

Two medians AD and BE of a ∆ABC meet each other at O. Prove that AO=OD

Answer 1

ANSWER

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)

ΔBFD∼ΔBEC (AA similarity)

∴

BE

BF

=

BC

BB

∴

BE

BF

= 2

1

(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:-

∴

EF

AE

=

GD

AG

........(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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