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**.