CAT The Mid-point Theorem - Practice Questions & MCQ

Let us study a result which is related to the mid-point of sides of a triangle. Before directly jump into the theorem, let's do an activity first.

Draw a triangle (say ABC) and mark the mid-points D and E of two sides of the triangle. Join the points D and E. Measure DE and BC. Measure ∠ ADE and ∠ ABC. What do you observe?

Theorem 9 : The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

Let's prove this

Given a ∆ ABC in which D and E are the midpoints of AB and AC respectively. DE is joined.

We have to prove, DE || BC and DE = (1/2)BC

Draw CF parallel to BA, meeting DE produced in F.

In ∆ AED and ∆ CEF, we have

                        ∠ AED = ∠ CEF            (vertically opposite angle)

                              AE = CE                 (E is the mid-point of AC)

                        ∠ DAE = ∠ FCE           (alternate interior angle)

Therefore,       ∆ AED ≅ ∆ CEF                (ASA criteria)

And so,       AD = CF and  DE = EF      (Corresponding Parts of Congruent Triangles)

But,                       AD = BD                   (D is the midpoint of AB)

and,                      BD || CF

Therefore,  BD = CF and BD || CF

This implies, BCFD is aparallelogram.

⇒               DF || BC and DF = BC

⇒               DE = BC and DE = (1/2)DF = (1/2)BC

Hence, DE || BC and DE = (1/2)BC.

The converse of Midpoint Theorem

Theorem 10 : The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.

Given a ∆ ABC in which D is the midpoints of AB amd DE is parallel to BC.  

We have to prove E is the mid-point of AC.

Draw CF parallel to BA, meeting DE produced in F.

We have

                          DF || BC                    (because DE || BC)

                          BD || CF                    (because CF || BA)

Therefore, DBCF is a parallelogram.

                      CF = DB = AD              (D is the mid point of AB)

Now, in ∆ ADE and ∆ CFE, we have

                     ∠ EAD = ∠ ECF            (alternate interior angle)

                          AD = CF

and              ∠ ADE = ∠ CFE             (alternate interior angle)

Therefore,   ∆ ADE ≅ ∆ CFE   (         ASA criteria)

and so,              AE = CE   (Corresponding Parts of Congruent Triangles)

Hence, E is the midpoint of AC.