Question

In the following figure, medians AD and BE of ΔABC meet at point G and DF || BE. Prove that: AG : GD = 2 : 1

Answer 1

Given:

In △ABC, AD and BE are medians, and they intersect at G.

Since BE is the median of △ABC, E is the midpoint of AC.

So,

AE = EC

Also, from DF || BE:

EF = FC

Hence, F is the midpoint of EC.

Now in △AEC, since F is the midpoint of EC and DF || AE, by the Midpoint Theorem, D is the midpoint of AC.

But D is already the midpoint of BC, since AD is a median.

Now consider △ABE and △DGF:

Since DF || BE,

∠GDF = ∠GAB, ∠GFD = ∠GBA, and ∠G = ∠G

Therefore,

△GDF∼△GBE

So, `(GD)/(GA) = (DF)/(BE)`

Now in △CBE, D is the midpoint of CB and DF || BE.

Therefore, by the Midpoint Theorem,

`DF = 1/2 BE`

Thus, `(GD)/(GA) = 1/2`

So,

GA = 2GD or AG : GD = 2 : 1

Hence proved.