Given: BD.BC = BA.BF

To prove: ∆BDC ~ ∆BAF

Proof:

From the given equation, we can rewrite it as BD/BA = BF/BC

Now, let's consider ∆BDC and ∆BAF:

By the Angle-Angle Similarity Theorem, if two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.

∠BDC = ∠BAF (Vertically opposite angles)
∠BCD = ∠BFA (Given)

Therefore, by the Angle-Angle Similarity Theorem, we can conclude that ∆BDC ~ ∆BAF.