Question

In a triangle, explain how can be justify that the sum of two sides is less than the sum of third side and twice the median to the third side.​

Answer 1

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Given : Triangle ABC in which AD isthe 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

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