Converse of midpoint theorem

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

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

Line MN is parallel to BC.
MN || BC

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

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.

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

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