Question

In the given figure, side BC of a ∆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

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 ∆ ABX, we get: 

AO/ AX = AF/ AB …(1) 

Also, in ∆ 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 ∆ ABC, EF || CB. 

This completes the proof.