# Midpoint theorem of triangle with proof

In this chapter we will discuss midpoint theorem of triangle along with the proof.

## What is midpoint theorem of triangle ?

It states that any line joining the mid point of two adjacent sides of triangle is parallel to third side and has half of the length.

In the above triangle ABC, M & N are midpoints of sides AB and AC respectively.

According to the midpoint theorem, the line MN is parallel to third side BC and is equal to half of BC.

\mathtt{MN\ =\frac{BC}{2}}

### Proof of midpoint theorem of triangle

Given:
Given below is the triangle ABC in which M & N are midpoints of side AB & AC respectively.

MN is the line segment joining the points M & N.

Construction
Extend line MN to point O such that line OC is parallel to AB.
i.e. OC || AB

Also, join points MC

To prove:
Prove that MN is parallel to BC and \mathtt{MN\ =\frac{BC}{2}}

Proof
Consider triangle ANM and CNO

∠ANM = ∠ CNO { Vertically opposite angle }
∠MAN = ∠NCO { interior angle }
AN = NC { N is midpoint of AC }

By AAS congruency condition, triangle ANM and CNO are congruent.
i.e. \mathtt{\triangle ANM\ \cong \triangle CNO}

Since both the triangles are congruent, we can say that;
AM = CO
MN = NO

Since AM = MB, we can say that; AM = MB = CO.

Now consider triangle MBC and COM

CM = MC { common side }
∠ BMC = ∠ MCO { alternate angles }
MB = OC { Proved above }

By SAS congruency condition, both triangles MBC and COM are congruent.
i.e. \mathtt{\triangle MBC\ \cong \triangle COM}

Since both the triangles are congruent we can say that;
BC = MO and ∠MCB = ∠ CMO

Taking BC = MO;

BC = MN + NO

We have already proved that MN = NO; the expression can be written as;
BC = MN + MN

BC = 2MN

\mathtt{MN\ =\frac{BC}{2}}

Hence, we proved that line MN is half of line BC.

Also since ∠MCB = ∠ CMO.
Both angles are interior angles, it means that line MN is parallel to BC.