Question

The side BC of a $\Delta$ 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 O is the mid-point of OX. Prove that AO : AX = AF : AB and show that FE $\parallel$ BC.

Answer 1

Join BX and CX

We have,

BD = CD and OD = DX

Thus, BC and OX bisect each other.

$\Rightarrow$ OB XC is a parallelogram.

$\Rightarrow$ BX II CO and CX II BO

$\Rightarrow$ BX II CF and CX II BE

$\Rightarrow$ BX II OF and CX II OE

In $△$ABX, we have

BX II OF

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

In $△$ACX, we have

CX II OE

$\Rightarrow$ $\frac{AO}{AX}$ = $\frac{AE}{AC}$

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

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

Thus, E and F are points on AB and AC such that they divide AB and AC respectively in the same ratio.

Therefore, by the converse of Thale's Theorem FE || BC.