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