Question

Prove that any two sides of a triangle
are together greater than twice the median drawn to​

Answer 1

Answer:

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Step-by-step explanation:

Given: In △ABC,AD is the median drawn from A to BC

To Prove that AB+AC>AD

Construction:Produce AD to E so that DE=AD, join BE

Proof:In △ADC and △EDB we have

AD=DE(constant)

DC=BD as D is the midpoint

∠ADC=∠EDB (vertically opposite angles)

∴ In △ABE, △ADC≅△EDB by S.A.S

This gives BE=AC

AB+BE>AE

AB+AC>2AD ∵AD=DE and BE=AC

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