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Sum

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)`.

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#### 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 = 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.

Concept: Basic Proportionality Theorem (Thales Theorem)

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