Question

In A ABC, AD is a median and O is any point on AD. B0 and CO on producing meet AC and AB at E and F
respectively. Now AD is produced to X such that OD = DX as shown in figure.
Prove that :
[4]
(1) EF|| BC
(2) AO: AX = AF: AB

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Answer 1

Answer:

Explanation:

BC and OX bisect each other So, BXCO is a parallelogram, BE || XC and BX || CF  

In ΔABX, by B.P.T ,

AF / FB = AO / OX .......................(1)

IN  Δ AXC

AE / EC = AO / OX ........................(2)

THUS, AF /  FB  =  AE  / EC

HENCE ,

FE ║BC ............................ (CONVERSE OF BPT)

NOW WE KNOW THAT

AF / FB =  AO / OX ..................(FROM (1))

ADDING  1 TO BOTH THE SIDES WE GET

AB / AF = AX / AO

BY RECIPROCATING WE GET

AO : AX = AF : AB