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 = GF   .....(i)

In ΔACG, CG || EF and E is mid-point of CA.

So, CE = GF  [By mid-point theorem]  .....(ii)

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.