Question

In a triangle ABC, let D be the mid-point of BC. Prove that AB + AC > 2AD. (What property of quadrilateralis needed here?)

Answer 1

i) This can be proved by making a small construction. Extend AD to E such that AD = DE. Join BE and CE. So, AE = 2AD ---------- (1) 

ii) In the quadrilateral, ABEC, the diagonal AE and BC bisect each other at D. [Since by construction AD = DE and as given AD is the median to BC; so D is the midpoint of BC] 
Since diagonals bisect each other, the quadrilateral ABEC is a parallelogram. 
In a parallelogram opposite sides are equal; so BE = AC --------- (2) 

iii) In triangle ABE, AB + BE > AE [Sum of any two sides of a triangle is greater than 
the third side] 

So from (1) & (2) 
AB + AC > 2AD
HOPE THIS WILL HELP. PLEASE MARK AS BRAINLIEST AND FOLLOW ME.

Answer 2

From the figure ,

i ) In a triangle ABC , AD is the median

drawn on the side BC is produced

to E such that AD = ED then ABCD

is a parallelogram .

AE = AD + DE

=> AE = 2AD ---( 1 )

AC = BC ---( 2 )

[ Opposite sides of parallelogram ]

ii ) The sum of any two sides of a

triangle is greater than the third side .

Now ,

In ∆ABE ,

AB + BE > AE

=> AB + AC > 2AD [ from ( 1 ) & ( 2 ) ]

•••••