In the given figure, side BC of $\triangle\text{ABC}$ 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.
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It is given that BC is bisected at D.
$\therefore\text{BD}=\text{DC}$
It is also given that OD = OX
The diagonals OX and BC of quadrilateral BOCX bisect each other.
Therefore, BOCX is a parallelogram
$\therefore\text{BO }||\text{ CX}$ and $\text{BX }||\text{ CO}$
$\text{BX }||\text{ CF}$ and $\text{CX }||\text{ BE}$
$\text{BX }||\text{ OF}$ and $\text{CX }||\text{ OE}$
Applying Thales' theorem in $\triangle\text{ABX},$ we get:
$\frac{\text{AO}}{\text{AX}}=\frac{\text{AF}}{\text{AB}}\dots(1)$
Also, in $\triangle\text{ACX},\text{CX }||\text{ OE}.$
Therefore by Thales' theorem, we get:
$\frac{\text{AO}}{\text{AX}}=\frac{\text{AE}}{\text{AC}}\dots(2)$
From (1) and (2), we have:
$\frac{\text{AF}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}$
Applying the converse of Thales' theorem in $\triangle\text{ABC},\text{EF }||\text{ CB}.$
This completes the proof.
$\therefore\text{BD}=\text{DC}$
It is also given that OD = OX
The diagonals OX and BC of quadrilateral BOCX bisect each other.
Therefore, BOCX is a parallelogram
$\therefore\text{BO }||\text{ CX}$ and $\text{BX }||\text{ CO}$
$\text{BX }||\text{ CF}$ and $\text{CX }||\text{ BE}$
$\text{BX }||\text{ OF}$ and $\text{CX }||\text{ OE}$
Applying Thales' theorem in $\triangle\text{ABX},$ we get:
$\frac{\text{AO}}{\text{AX}}=\frac{\text{AF}}{\text{AB}}\dots(1)$
Also, in $\triangle\text{ACX},\text{CX }||\text{ OE}.$
Therefore by Thales' theorem, we get:
$\frac{\text{AO}}{\text{AX}}=\frac{\text{AE}}{\text{AC}}\dots(2)$
From (1) and (2), we have:
$\frac{\text{AF}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}$
Applying the converse of Thales' theorem in $\triangle\text{ABC},\text{EF }||\text{ CB}.$
This completes the proof.
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