Question

In the given figure, side BC of ∆ABC is bisected at D and O is any point on AD. BO and CO produced meet AC and AB at E and F, respectively, and AD is produced to X, so that D is the midpoint of OX. Prove that AO : AX = AF : AB and show that EF ∥ BC.

Answer 1

Join BX and CX.
It is given that BC is bisected at D.
∴ BD = DC ​
It is also given that OD = OX
The diagonals OX and BC of quadrilateral BOCX bisect each other.
Therefore, BOCX is a parallelogram.
∴ BO $\parallel CX\hspace{0.17em}\mathrm{and}BX\parallel CO$
$BX\parallel CF\mathrm{and}CX\parallel BE$
$BX\parallel OF$ and $CX\parallel OE$
Applying Thales' theorem in $∆$ABX, we get:
$$\begin{gathered} \frac{AO}{AX}=\frac{AF}{AB}...\left(1\right) \\ \mathrm{Also},\mathrm{in}△ACX,CX\parallel OE. \\ \mathrm{Therefore}\mathrm{by}\mathrm{Thales}\text{'}\mathrm{theorem},\mathrm{we}\mathrm{get}: \\ \frac{AO}{AX}=\frac{AE}{AC}...\left(2\right) \end{gathered}$$
From (1) and (2), we have:
$\frac{AF}{AB}=\frac{AE}{AC}$
Applying the converse of Thales' theorem in $△ABC,EF\parallel CB$.
This completes the proof.