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 $\mathrm{\Delta }CDE$ and $\mathrm{\Delta }EBF,$

$CE=BE$ ; [Given]

$DE=EF$ ; [According to the construction]

$\mathrm{\angle }CED=\mathrm{\angle }AEF$ ; [Vertical Angles]

∴ $\mathrm{\Delta }CDE\cong \mathrm{\Delta }EBF$ ; [Side-Angle-Side theorem]

So, $\mathrm{\angle }CDE=\mathrm{\angle }BFE$

∴ $CD\parallel BF$

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, $\mathrm{\Delta }CDE\cong \mathrm{\Delta }EBF$]

Or, $2DE=AB$

Or, $DE=\frac{1}{2}AB$

∴ In the triangle $\mathrm{\Delta }ABC,$ $DE\parallel AB$ 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 $\mathrm{\Delta }ABC,$ $DE\parallel BC$ and $DE=\frac{1}{2}BC$, where $D$ and $E$ are the midpoints of $AC$ and $AB$. [Proved]