# Timeless Theorems of Mathematics/Mid Point Theorem

!In this right triangle, $AD = CD, BE = CE, DE = \frac {1}{2}AB$and
$DE \parallel AB$ according to the 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

!The construction for the mid-point theorem\'s proof with similar
triangles
**Proposition:** Let $D$ and $E$ be the midpoints of $AC$ and $BC$ in
the triangle $ABC$. It is to be proved that,

1.  $DE \parallel AB$ and;
2.  $DE = \frac {1}{2}AB$.

**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,

1.  $DE = \frac {1}{2}BC$ and
2.  $DE \parallel BC$

**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
$AB$. ***\[Proved\]***
