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Question

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

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Solution

Join BX and CX

We have,

BD = CD and OD = DX

Thus, BC and OX bisect each other.

OB XC is a parallelogram.

BX II CO and CX II BO

BX II CF and CX II BE

BX II OF and CX II OE

In ABX, we have

BX II OF

AOAX = AFAB ..... (i)

In ACX, we have

CX II OE

AOAX = AEAC

From equations (i), (ii), we get

AFAB = AEAC

Thus, E and F are points on AB and AC such that they divide AB and AC respectively in the same ratio.

Therefore, by the converse of Thale's Theorem FE || BC.


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