Question

If a line drawn parallel to one side of a triangle intersecting the other two sides in distinct points divides the two sides in the same ratio, then it is parallel to the third side.

State and prove the converse of the above statement.

Answer 1

Given:

In △ABC, a line DE intersects AB at D and AC at E, such that

`"AD"/"DB" = "AE"/"EC"`

To Prove:

DE ∥ BC

Assume that DE is not parallel to BC. Draw a line D′E′ parallel to BC that intersects AB at D′ and AC at E′.

Proof:

Since D′E′ ∥ BC, by the Basic Proportionality Theorem,

`"AD'"/"D'B" = "AE'"/"EC"`

`"AD"/"DB" = "AD'"/"D'B"`

Since the ratios are equal and the points D and D′ lie on the same segment AB, it follows that D and D′ must coincide. Similarly, E and E′ coincide.

Thus, DE and D′E′ are the same line, meaning DE ∥ BC.

Thus, if a line divides two sides of a triangle in the same ratio, it must be parallel to the third side. Hence, the converse of the Basic Proportionality Theorem is proved.