In this chapter we will discuss the converse of midpoint theorem and its proof.

## What is midpoint theorem converse ?

It states that in a triangle, the line joining the midpoint of one side and parallel to another side will definitely bisect the third side.

For example, consider the above triangle ABC.

The **point M is the midpoint of side AB** and **segment MN is parallel to BC**.

According to converse of midpoint theorem, **the line segment MN formed by joining midpoint M of side AB and is parallel to side BC will bisect the side AC**.

i.e. AN = NC

### Proof of midpoint theorem converse

**Given:**

ABC is a triangle in which M is midpoint of AB.**AM = MB**.

Line MN is parallel to BC.**MN || BC**

**Construction:**

Draw line CO parallel to line AB.**CO || AB**

Extend line MN to meet line CO at P.**To prove:**

N is the midpoint of AC**Proof**

Consider the **quadrilateral MBCP**.

Here MB is parallel to CP

MB || CP.

Also, MN parallel to BC. { given}

MN || BC

Since opposite sides are parallel to each other, **the quadrilateral MBCP is a parallelogram**.

We know that in parallelogram, **opposite sides are equal.**

So, MB = CP and MP = BC.

Since, M is the midpoint of AB.

AM = MB

We can write, **AM = MB = CP.**

Now consider **triangle AMN and CPN.**

∠ANM = ∠CNP {vertically opposite angle }

∠MAN = ∠NCP {alternate angle since line AB & CO are parallel }

AM = CP { proved above }

By **AAS congruency condition**, the triangle AMN and CPN are congruent.

\mathtt{\triangle AMN\ \cong \triangle CPN}

Since both triangles are congruent we can say that, AN = NC.

So, **N is the midpoint of side AC.**

Hence Proved