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Standard X

Mathematics

Basic Proportionality Theorem

In the given ...

Question

In the given figure, side BC of $Δ A B C$ is bisected at D and O is any point on AD, BO and CO produced meet AC and AB at E and F respectively, and AD is produced to X so that D is the midpoint of OX. Prove that AO : AX = AF : AB and show that EF || BC.

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Solution

Given BD = CD and OD = DX

Join BX and CX

Thus, the diagonals of quad OBXC bisect each other

OBXC is a parallelogram

BX || CF and so, OF || BX

Similarly, CX || OE

In ΔABX, OF || BX

Therefore,
AO/AX=AF/AB—– (1)

In ΔACX, OE || XC

Therefore,
AO/AX=AE/AC—– (2)

From (1) & (2) we get:

AF/AB=AE/AC

Hence, EF || BC

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