Question

In the given figure, side BC of ΔABC 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

It is give 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 || CX and BX || CO
BX || CF and CX || BE
BX || OF and CX || OE 

Applying Thales’ theorem in Δ ABX, we get: 

`(AO)/(AX)=(AF)/(AB)`           ............(1) 

Also, in Δ ACX, CX || OE.
Therefore by Thales’ theorem, we get: 

`(AO)/(AX)=(AE)/(AC)`   ..................(2) 

From (1) and (2), we have: 

`(AO)/(AX)=(AE)/(AC)` 

Applying the converse of Theorem in Δ ABC, EF || CB.
This completes the proof.