Parallelograms Theorem 7

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Parallel lines and Triangles

So far we have proved various theorems on parallelograms. Let us now apply these theorems to prove a few interesting and useful facts about a triangle.

Statement

"The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it".

Given:

In D ABC, AD=DB and AE=EC

To prove:

Construction:

Analysis for construction shows Analysis for construction: that you have to draw CF||BD Think how you can complete to meet DE produced at F. a parallelogram with DB and BC as consecutive sides. You will find the need to draw CF||DB.

Proof:

In triangles, ADE and CEF,

AE=EC (given)

AD=CF and DE=EF (corresponding parts of corresponding triangles)

But AD=DB (given)

DB=CF ----(i)

(AD is equal to both DB and CF)

In quadrilateral DBCF,

DB=CF and DB||CF

DBCF is a parallelogram. (by definition of parallelogram)

DF=BC

(opposite sides of a parallelogram are equal)

and DF||BC ----(ii)

But DE = EF (proved above)

And DF=DE+EF

=2 DE

and DF=BC (from (ii))

BC=2 DE

Hence the theorem is proved.