In Δ ABC, the AD is a median and E is the midpoint of the AD. If BE is produced to meet 1 AC in F, show that AF = 1/3 AC.
Proof:
Given: In ΔABC, the AD is a median and E is mid-point of the AD and BE is produced to meet AC in F.
To Prove:
Construction: Draw DG || BF as shown in figure
Proof:
Now, In ΔBFC
DG || BF [By construction]
As AD is a median on BC, D is a mid-point of BC
Therefore,
G is a mid-point of CF [By mid-point theorem]
⇒ CG = FG …[1]
Now, In ΔADG
EF || DG [By Construction]
As E is a mid-point of AD [Given]
Therefore,
F is a mid-point of AG [By mid-point theorem]
⇒ FG = AF …[2]
From [1] and [2]
AF = CG = FG …[3]
And
AC = AF + FG + CG
⇒ AC = AF + AF + AF [From 3]
⇒ AC = 3AF
Hence Proved