The median AD of the triangle ABC is bisected at E and BE meets AC at F. Find AF:FC.

The median AD of the triangle ABC is bisected at E. BE meets AC in F then AF : AC is

The median AD of triangle ABC is bisected at F and BF is produced to meet the side AC in P. If AP=lambda AC then what is the value of lambda

The median AD of a triangle ABC is bisected at F, and BF is produced to meet the side AC in P. If AP = lamdaAC , then what is the value of lamda .?

A median drawn from the vertex from the A of a triangle ABC is bisected at E . BE meets AC in F such that AF:AC = 1 : n where n is-

The median AD to DeltaABC is bisected at E and BE is produced to meet the side AC in F. Show that bar(AF)=1/3bar(AC)

the median AB of Delta ABC is bisected at Eand BE is produced to meet the side AC in f.then the ratio AF:FC=

In the fig. AD is a median of triangle ABC and E is the mid point of AD, also BE meets AC in F. prove that AF=1/3AC

The side BC of a triangle ABC is bisected at D; O is any point in AD. BO and CO produced meet AC and AB in E and F respectively and AD is produced to X so that D is the mid-point of OX. Prove that AO:AX=AF:AB and show that FE||BC.