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

Given: ΔABC in which AD is a median.

To prove: AB + AC > 2AD.

Construction: Produce AD to E, such that AD = DE. Join EC.

BD = BD (D is the mid point of BC)

∠ADB = ∠EDC (Vertically opposite angles)

∴ ΔADB  ΔEDC (SAS congruence criterion)

⇒ AB = ED (CPCT)

In ΔAEC,

AC + ED > AE (Sum of any two sides of a triangles is greater than the third side)

Cheers!

• 140

itzz pythogoras theorem

• -13

pythagoras

• -16

thanq ...

• -12

suppose if we take all the angles as x+x+x=180

so x+x=2x which therefore is greater then x

2x>x

easy question.

• -34

Given: ΔABC in which AD is a median.

To prove: AB + AC > 2AD.

Construction: Produce AD to E, such that AD = DE. Join EC.

BD = BD (D is the mid point of BC)

∠ADB = ∠EDC (Vertically opposite angles)

∴ ΔADB  ΔEDC (SAS congruence criterion)

⇒ AB = ED (CPCT)

In ΔAEC,

AC + ED > AE (Sum of any two sides of a triangles is greater than the third side)

Cheers!

• 9

same question appered in my maths paper........

• -13

umangbansal600, u were a bit wrong. It is AB=EC, not ED. Thus it is AC+EC>AE. Therfore, AC+AB>2AD

• 55

HOW IS IT PYTHOGORAS????????

• -26

Anshudude is right!

• -11

How its a pythogoras????? because in this theorem  Hypotaneouse of a rightangle Triangle is equal to the sum of remaining two sides.

• -5
What are you looking for?