Question

AD is a median of side BC of ABC. E is the midpoint of AD. BE is joined and produced to meet AC at F. Prove that AF: AC = 1 : 3.

Answer 1

Construction: Draw DS ∥ BF, meeting AC at S.

Proof: 
In ΔBCF, D is the mid-point of AC DS || BF.
∴ S is the mid-point of CF.
⇒ CS = SF      ....(i)
In ΔADS, E is the mid-point of AD and EF || DS.
∴ F is the mid-point of AS.
⇒ AF = FS      ....(ii)
From (i) and (ii), we get
AF = FS = SC
⇒ AC = AF = FS + SC
⇒ AC = AF + AF +AF
⇒ AC = 3AF
⇒ `"AF"/"AC" = (1)/(3)`
⇒ AF : AC = 1 : 3.