Question

In the following figure, DE || AC and `(BE)/(EC) = (BC)/(CF)`, prove that DC || AF.

Answer 1

Given:

• DE || AC
• `(BE)/(EC) = (BC)/(CF)`

In △BAC, since DE∥AC, by the Basic Proportionality Theorem,

`(BD)/(DA) = (BE)/(EC)`

But we have given as: `(BE)/(EC) = (BC)/(CF)`

So, `(BD)/(DA) = (BC)/(CF)`

Now B, C, F are collinear, and B, D, A are collinear.

From `(BD)/(DA) = (BC)/(CF)`, the corresponding sides around the angle at D and C are proportional.

Thus, by the converse of the Basic Proportionality idea, the line through D and C is parallel to AF.

Therefore,

DC || AF

Hence proved.