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

Question

Philoid · Accepted Answer

We have the diagram as,

Given: BD = DC & OD = DX

To Prove: and also, EF ∥ BC

Proof: Since, from the diagram we can see that diagonals OX and BC bisect each other in quadrilateral BOCX. Thus, BOCX is a parallelogram.

If BOCX is a parallelogram, BX ∥ OC, and BO ∥ CX.

⇒ BX ∥ FC (as OC extends to FC) and CX ∥ BE (BO extends to BE)

⇒ BX ∥ OF and CX ∥ OE

∵ BX ∥ OF, applying Thale’s theorem in ∆ABX, we get

…(i)

Now since CX ∥ OE, applying Thale’s theorem in ∆ACX, we get

…(ii)

By equations (i) and (ii), we get

By applying converse of Thale’s theorem in the above equation, we can write

EF ∥ BC

Hence, proved.