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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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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.
HOPE IT WILL HELP YOU MARK ME BRAINLIEST
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