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Question
In the following figure, AD and CE are medians of ΔABC. DF is drawn parallel to CE. Prove that :
- EF = FB,
- AG : GD = 2 : 1
Solution
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
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