Question

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.

Answer 1

According to the question,

$BD=CD$ and $OD=DX$

Therefore, BC and OX bisect each other and OBXC is a parallelogram.

This gives $BX\parallel CO$ and $CX\parallel BO$

or $BX\parallel CF$ and $CX\parallel BE$

or $BX\parallel OF$ and $CX\parallel OE$

In $\Delta ABX$, as $BX\parallel OF$, then,

$\frac{AO}{AX}=\frac{AF}{AB}$ .............(1)

In $\Delta ACX$, as $CX\parallel OE$, then,

$\frac{AO}{AX}=\frac{AE}{AC}$ ............. (2)

From equation (1) and (2),

$\frac{AF}{AB}=\frac{AE}{AC}$

Hence, as E and F divides AB and AC respectively in the same ratio, so, $FE\parallel BC$.