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The side B C of a triangle A B C is bise...

The side `B C` of a triangle `A B C` is bisected at `D ; O` is any point in `A D. BO and CO` produced meet `AC and AB` in `E and F` respectively and `A D` is produced to `X` so that `D` is the mid-point of `O X`. Prove that `A O : A X=A F : A B` and show that `FE||BC`.

Text Solution

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Join BX and CX.
Since D is the mid -point of BC ( given )
BD= DC …(1)
Also D is the mid-point of OX (given )
OD= DX … (2)
From (1) and (2) we, have
`squareOBXC` is a parallelogram ( diagonals bisect each other)
OC||BX ( opposite sides of a parallelogram)
`Rightarrow FC||BX Rightarrow FO ||BX`
`In DeltaABX, (AO)/(OX)= (AF)/(FB)` ( by B.P theorem) ...(3)
Also , BO||XC ( opposite sides of a parallelogram)
` Rightarrow BE||XCRightarrow OE||XC`
` (AO)/(OX)= (AE)/(EC)` (by B.P theorem) ...(4)
From (3) and (4) , we have
`1 ( AF)/(FB) = (AE) (EC)`
In ` DeltaABC, FE||BC.` (converse of B.P theorem) Hence Proved.
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