Question

. In the figure, seg DE || side AC and
seg EF || side BA then prove that
AD AF = 1
DB FC​

Answer 1

Attached below, hope it helps

Answer 2

In the given figure triangles ABC and DBE are similar and triangles ABC and CEF are similar.

▪From triangle ABC and triangle DBE, as AC is parallel to DE, BD/AD = BE/EC -- (1)

▪From triangle ABC and triangle CEF, as AB is parallel to EF, CF/AF = EC/BE => AF/CF = EC/BE -- (2)

▪From equations 1 and 2 we have

BE/EC = BD/AD = AF/CF

=> BD/AD = AF/CF

=> (AD/BD)×(AF/CF) =1