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Question
In the adjoining figure, the medians BD and CE of a ∆ABC meet at G. Prove that
(i) ∆EGD ∼ ∆CGB and
(ii) BG = 2GD for (i) above.
Solution
Since D and E are mid-point of AC and AB respectively in ∆ABC, Ed is parallel to BC.
(i) In ∆'s EGD and CGB,
∠EGD = ∠CGB ...(Vertically opp., angles)
∠EGD = ∠CBG ...(Alternative angles)
So, ∆EGD ∼ ∆CGB.
Hence proved.
(ii) ∴ `"BG"/"GD" = "BC"/"DE"`
But `"BC"/"DE" = 2, "So" "BG"/"GD" = 2`
⇒ BG = 2GD.
Hence Proved.
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