Q. 173.9( 9 Votes )

# The side BC of a triangle ABC is bisected at D; O is any point in AD. BO, CO produced meet AC, AB in E, F respectively, and AD is produced to X so that D is the mid point of OX. Prove that AO : AX = AF : AB and show that EF is parallel to BC.

Answer :

**Given :**The side BC of a triangle ABC is bisected at D; O is any point in AD. BO , CO produced meet AC, AB in E , F respectively, and AD is produced to X so that D is the mid point of OX.

**To Prove :**AO : AX = AF : AB and show that EF is parallel to BC.

**Construction:**Join BX and CX.

**Proof:**In quadrilateral BOCX, BD = DC and DO = DX (given)

∴ BOCX is a parallelogram (When the diagonals of a quadrilateral bisect each other, then the quad. is a parallelogram)

∴ BX || CO (Definition of a parallelogram)

or BX || FO.

In Δ ABX, BX || FO(proved).

∴ AO : AX = AF : AB (using B.P.T) -----------(i)

Similarly, AO : AX = AE : AC --------------(ii)

From (i) and (ii),

AF : AB = AE : AC

By corollary to B.P.T, EF is parallel to BC.

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