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.

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, Construction: Add $$D$$ and $$E$$, extend $$DE$$ to $$F$$ as $$EF = DE$$, and add $$B$$ and $$F$$.
 * 1) $$DE \parallel AB$$ and;
 * 2) $$DE = \frac {1}{2}AB$$.

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]

&there4; $$\Delta CDE \cong \Delta EBF$$ ; [Side-Angle-Side theorem]

So, $$\angle CDE = \angle BFE$$

&there4; $$CD \parallel BF$$

Or, $$AD \parallel BF$$ and $$CD = BF = DA$$

Therefore, $$ADFB$$ is a parallelogram.

&there4; $$DF \parallel AB$$ 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, $$DE = \frac {1}{2}AB$$

&there4; 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, Proof: [1] The distance of the segment $$BC = \sqrt{(x_3 - x_2)^2 + (y_3 - y_2)^2}$$
 * 1) $$DE = \frac {1}{2}BC$$ and
 * 2) $$DE \parallel BC$$

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 $$E(\frac {x_1 + x_2}{2}, \frac {y_1 + y_2}{2})$$

&there4; 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, $$DE \parallel BC$$

&there4; 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]