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Question
In the figure, given below, the medians BD and CE of a triangle ABC meet at G. Prove that:
- ΔEGD ~ ΔCGB and
- BG = 2GD from (i) above.
Solution
i. Since, BD and CE are medians
AD = DC
AE = BE
Hence, by converse of Basic Proportionality theorem,
ED || BC
In ΔEGD and ΔCGB,
∠DEG = ∠GCB ...(Alternate angles)
∠EGD = ∠BGC ...(Vertically opposite angles)
ΔEGD ~ ΔCGB ...(AA similarity)
ii. Since, ΔEGD ~ ΔCGB
`(GD)/(GB) = (ED)/(BC)` ...(1)
In ΔAED and ΔABC,
∠AED = ∠ABC ...(Corresponding angles)
∠EAD = ∠BAC ...(Common)
∴ ΔEAD ∼ ΔBAC ...(AA similarity)
∴ `(ED)/(BC) = (AE)/(AB) = 1/2 ` ...(Since, E is the mid-point of AB)
`=> (ED)/(BC) = 1/2`
From (1),
`(GD)/(GB) = 1/2`
GB = 2GD
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