D is the midpoint of side BC of ΔABC. Point E lies on AC such that CE = $\frac{1}{3}$AC. BE and AD intersect at G. What is $\frac{AG}{GD}$ ?

4 : 1

Given,

D is the midpoint of side BC

So, BD : DC = 1 : 1

Point E lies on AC such that CE = \(\frac{AC}{3}\)

So, AC : CE = 3 : 1

So, CE is 1 and AE = AC - CE

So, AE = 2

AE : CE = 2 : 1

From the following figure balancing sides of the triangle,

LCM of 2 ,1,1 and 1 is 2

Therefore, AG : GD = 4 : 1