Question

In the following figure, AD and CE are medians of ΔABC. DF is drawn parallel to CE. Prove that :

1. EF = FB,
2. AG : GD = 2 : 1

Answer 1

i. In ∆BFD and ∆BEC,

∠BFD = ∠BEC   ...(Corresponding angles)

∠FBD = ∠EBC   ...(Common)

∆BFD ~ ∆BEC   ...(AA Similarity)

∴ `(BF)/(BE) = (BD)/(BC)`

`(BF)/(BE) = 1/2`  ...(As D is the mid-point of BC)

BE = 2BF

BF = FE = 2BF

Hence, EF = FB

ii. In ΔAFD, EG || FD.

Using Basic Proportionality theorem,

`(AE)/(EF) = (AG)/(GD)`  ...(1)

Now, AE = EB   ...(As E is the mid-point of AB)

AE = 2EF  ...(Since, EF = FB, by (i))

From (1),

`(AG)/(GD) = 2/1`

Hence, AG : GD = 2 : 1