The median AD, BE and CF in a ΔABC intersect at point G. Prove that

[Hint :AG + BG > AB]

As we know that in a triangle, sum of any two sides is always greater than the third side.

So, in ∆AGB,

AG+GB>AB…………….(1)

Also, AD=AG+GD

And BE=BG+GE

As we know that if the medians in a triangle intersect at a point then the point is the centroid of the triangle and the centroid divides the median in 2:1 ratio.

Here, G is the centroid of the ∆ABC.

So, AG:GD=2:1 and BG:GE=2:1

⇒ AD=3GD…………………..(2)

And AG=2GD…………………(3)

Dividing equation (2) by (3)

⇒

……………….(4)

Again,

⇒ BE=3GE…………………..(2)

And BG=2GE…………………(3)

Dividing equation (2) by (3)

⇒

……………….(5)

Putting the value of AG and BG in equation (1)

AG+BG>AB

⇒

⇒

Proved.