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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​

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 AXCAE / EC = AO / OX ........................(2)THUS, AF /  FB  =  AE  / ECHENCE ,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



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