# Timeless Theorems of Mathematics/Mid Point Theorem

The midpoint theorem is a fundamental concept in geometry that establishes a relationship between the midpoints of a triangle's sides. This theorem states that when you connect the midpoints of two sides of a triangle, the resulting line segment is parallel to the third side. Additionally, this line segment is precisely half the length of the third side.

## Proof
### Statement

In a triangle, if a line segment connects the midpoints of two sides, then this line segment is parallel to the third side and half its length.

### Proof with the help of Congruent Triangles

Proposition: Let $D$ and $E$ be the midpoints of $AC$ and $BC$ in the triangle $ABC$. It is to be proved that,
* $DE\parallel AB$ and;
* $DE={\frac {1}{2}}AB$.
Construction: Add $D$ and $E$, extend $DE$ to $F$ as $EF=DE$, and add $B$ and $F$.
Proof: [1] In the triangles $\Delta CDE$ and $\Delta EBF,$
$CE=BE$ ; [Given]
$DE=EF$ ; [According to the construction]
$\angle CED=\angle AEF$ ; [Vertical Angles]
∴ $\Delta CDE\cong \Delta EBF$ ; [Side-Angle-Side theorem]
So, $\angle CDE=\angle BFE$
∴ $CD\parallel BF$
Or, $AD\parallel BF$ and $CD=BF=DA$
Therefore, $ADFB$ is a parallelogram.
∴ $DF\parallel AB$ or $DE\parallel AB$
[2] $DF=AB$
Or $DE+EF=AB$
Or, $DE+DE=AB$ [As, $\Delta CDE\cong \Delta EBF$]
Or, $2DE=AB$
Or, $DE={\frac {1}{2}}AB$
∴ In the triangle $\Delta ABC,$ $DE\parallel AB$ and $DE={\frac {1}{2}}AB$, where $D$ and $E$ are the midpoints of $AC$ and $BC$. [Proved]

### Proof with the help of Coordinate Geometry

Proposition: Let $D$ and $E$ be the midpoints of $AC$ and $AB$ in the triangle $ABC$, where the coordinates of $A,B,C$ are $A(x_{1},y_{1}),B(x_{2},y_{2}),C(x_{3},y_{3})$. It is to be proved that,
* $DE={\frac {1}{2}}BC$ and
* $DE\parallel BC$
Proof: [1] The distance of the segment $BC={\sqrt {(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}}$
The midpoint of $A(x_{1},y_{1})$ and $C(x_{3},y_{3})$ is $D({\frac {x_{1}+x_{3}}{2}},{\frac {y_{1}+y_{3}}{2}})$.
In the same way, The midpoint of $A(x_{1},y_{1})$ and $B(x_{2},y_{2})$ is $E({\frac {x_{1}+x_{2}}{2}},{\frac {y_{1}+y_{2}}{2}})$
∴ The distance of $DE={\sqrt {({\frac {x_{1}+x_{3}}{2}}-{\frac {x_{1}+x_{2}}{2}})^{2}+({\frac {y_{1}+y_{3}}{2}}-{\frac {y_{1}+y_{2}}{2}})^{2}}}$
$={\sqrt {({\frac {x_{1}+x_{3}-x_{1}-x_{2}}{2}})^{2}+({\frac {y_{1}+y_{3}-y_{1}-y_{2}}{2}})^{2}}}$
$={\sqrt {{\frac {(x_{3}-x_{2})^{2}}{4}}+{\frac {(y_{3}-y_{2})^{2}}{4}}}}$
$={\sqrt {\frac {(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}{4}}}$
$={\frac {\sqrt {(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}}{2}}$
$={\frac {1}{2}}BC$ ; [As, $BC={\sqrt {(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}}$]
[2] The slope of $BC,$ $m_{1}={\frac {y_{2}-y_{3}}{x_{2}-x_{3}}}$
The slope of $DE,$ $m_{2}={\frac {{\frac {y_{1}+y_{2}}{2}}-{\frac {y_{1}+y_{3}}{2}}}{{\frac {x_{1}+x_{2}}{2}}-{\frac {x_{1}+x_{3}}{2}}}}$ $={\frac {\frac {y_{1}+y_{2}-y_{1}-y_{3}}{2}}{\frac {x_{1}+x_{2}-x_{1}-x_{3}}{2}}}$ $={\frac {\frac {y_{2}-y_{3}}{2}}{\frac {x_{2}-x_{3}}{2}}}$ $={\frac {y_{2}-y_{3}}{x_{2}-x_{3}}}$ $=m_{1}$ ; [As, $m_{1}={\frac {y_{2}-y_{3}}{x_{2}-x_{3}}}$]
Therefore, $DE\parallel BC$
∴ In the triangle $\Delta ABC,$ $DE\parallel BC$ and $DE={\frac {1}{2}}BC$, where $D$ and $E$ are the midpoints of $AC$ and $AB$. [Proved]