Question

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

Answer 1

Given : Triangle ABC in which AD is the median.
To prove:AB+AC>2AD
Construction :

Extend AD to E such that AD=DE .
 Now join EC.

Proof:
In ΔADB and ΔEDC
AD=DE[ By construction]
D is the midpoint BC.[DB=DB]

ΔADB=ΔEDC [vertically opposite angles]

Therefore Δ ADB ≅  ΔEDC [ By SAS congruence criterion.]
--> AB=ED[Corresponding parts of congruent triangles ]
In ΔAEC,
AC+ED> AE [sum of any two sides of a triangle is greater than the third side]
AC+AB>2AD[AE=AD+DE=AD+AD=2AD and ED=AB]

Hence proved 

Answer 2

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