Question

Please solve this question.

Answer 1

: From Point D, draw DG parallelBF

In ΔADG, E is the mid-point of AD and EF||DG.

∴F is the mid point of AG  [Converse of the mid point theorem]

⇒ AF = FG  ... (i)

In ΔBCF, D is the mid point of BC and DG||BF

∴ G is the mid point of CF

⇒ FG = GC  ... (ii)

From (i) and (ii), we get,

AF = FG = GC  ... (iii)

Now, AF + FG + GC = AC

⇒ AF + AF + AF = AC  [Using (iii)]

⇒ 3AF = AC

ac=1/3 af