Q. 44.1( 9 Votes )

Prove that the su

Answer :


Given a ∆ABC in which AD, BE and CF are the medians on the sides BC , AC and AB.


To prove: AD+BE>CF


BE+CF>AD


AD+CF>BE


Proof:


We will extend AD to H to form ∆BHC


Also, AG=GH………………(1)


F is the midpoint of AB and G is the midpoint of AH.


So, by midpoint theorem,


FG||BH


And FG= 1/2 BH


Similarly, GC||BH and BG||CH


So, we can see from the above that


BGCH is a parallelogram.


So, BH=GC …………(2)


And BG=HC…………..(3)


Now, in ∆BGH,


BG+GH>BH


BG+AG>GC (from (1),(2))


So, BE+AD>CF


Similarly, BE+CF>AD


AD+CF>BE


Proved.

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