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Question
In figure, line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and ∠AEF = ∠AFE. Prove that `(BD)/(CD) = (BF)/(CE)`.
Solution
Given ΔABC, E is the mid-point of CA and ∠AEF = ∠AFE
To prove: `("BD")/("CD") = ("BF")/("CE")`
Construction: Take a point G on AB such that CG || EF
Proof: Since, E is the mid-point of CA
∴ CE = AE ...(i)
In ΔACG,
CG || EF and E is mid-point of CA
So, CE = GF ...(ii) [By mid-point theorem]
Now, In ΔBCG and ΔBDF,
CG || EF
∴ `("BC")/("CD") = ("BG")/("GF")` ...[By basic proportionality theorem]
⇒ `("BC")/("CD") = ("BF" - "GF")/("GF")`
⇒ `("BC")/("CD") = ("BF")/("GF") - 1`
⇒ `("BC")/("CD") + 1 = ("BF")/("CE")` ...[From equation (ii)]
⇒ `("BC" + "CD")/("CD") = ("BF")/("CE")`
⇒ `("BD")/("CD") = ("BF")/("CE")`
Hence proved.
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