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)`.

Answer 1

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.