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Prove that sum of any two sides of a tri...

Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.

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Given : `DeltaABC` with median AD.
To prove : `AB+AC gt 2AD`
Construction : Produce AD to E such that DE=AD and join EC.
Proof : In `DeltaADB` and `DeltaEDC`.
`:' {(AD=ED,("by construction")),(angle1=angle2,("vertically opposite angles")),(DB=DC,("given")):}`
`:. DeltaADB cong DeltaEDC` (By SAS criterion)
`:. AB=EC` (c.p.c.t.)
and `angle3=angle4` (c.p.c.t)
Now, in `DeltaAEC`, we have
`AC+CE gt AE` (`:'` sum of any sides of a triangle is greater than the third side)
`implies" "AC+CE gt AD+DE`
`implies" "AC+CE gt AD+AD" "( :' AD=DE)`
`implies" "AC+CE gt 2AD`
`implies" "AC+AB gt 2AD" "( :' AB=EC)`
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