In the given figure, side BC of a ∆ABC is bisected at D and O is any

1.	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» 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.

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»

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.