Question

prove that sum of two sides of a triangle is greater than twice the median respective to the third side

Answer 1

Here's your answer..

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Solution:-

Let ABC a triangle and AD a median.

First,produce the line AD to E such that
AD =DE.

In ∆ADB and ∆BDC:-
=>AD=ED
=>BD=BC
=> <ADB =<EDC (Vertically opposite)

So,
=>∆ADB congruent ∆ECD
Hence,
AB = CE (By c.p.c.t)

Now,
In ∆AEC,
=>AC+EC >AE
(Sum of two side of triangle id greater than the third side)

So,
We can say that ,
=>AB +AC >AE
(Since,AB =CE so we can write AC at place of EC)

=>AB +AC >AE
=>AB +AC >AD +DE
=>AB+AC >AD +AD(Since,AD=DE)
=>AB+AC >2AD

Hence,
It is proved that Sum of any two side of a triangle is greater than twice the median drawn to the third side.

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#See the attachment also..

Hope it helps you...