Q10 of 186 Page 184

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.

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.


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