Parallelograms Converse of Theorem 7

Statement

A straight line drawn through the mid-point of one side and parallel to another side of a triangle bisects the third side.

Given:

ABC in which D is the mid-point of AB and DE||BC.

To prove:

E is the mid-point of AC. i.e., to prove AE=EC.

Construction:

Since DE||BC, you can complete a parallelogram with DB and BC as consecutive sides.

Hence draw EF||BD to meet DE produced at F.

Proof:

In quadrilateral DBCF,

DB||CF (by construction)

DF||BC (given)

DBCF is a parallelogram.

DB=CF ----(i)

(opposite sides of a parallelogram)

But DB=AD ----(ii) (given)

From (i) and (ii), AD=CF

Now compare triangles, AED and CEF,

AD = CF

\ AE = EC (CPCT)

That is E is the mid point of AC.

Hence the theorem is proved.

Recall the above theorem and apply it to the diagram given.

In the diagram if D is the mid-point of AB and DE is drawn parallel to BC, then E Will be the midpoint of AC i.e., AE=EC.

Now if AX is drawn parallel to BC, then also AE=EC if AD=DB.

Now draw three parallel lines AB, CD, EF as shown in the figure.

Draw a transversal t₁ such that AB=BC.

Now draw another transversal t₂.

Measure DE and EF. You will find that DE=EF and AB=BC.

In the diagram, AD, BE and CF are three parallel lines.

AB and BC are equal intercepts made on t₁.

If any transversal is drawn, the intercepts made on it will also be equal.