View source for Timeless Theorems of Mathematics/Mid Point Theorem

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[[File:Midpoint theorem.svg|thumb|In this right  triangle, <math>AD = CD, BE = CE, DE = \frac {1}{2}AB </math>and <math>DE \parallel AB</math> 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 ===
[[File:Midpoint-theorem-construction.svg|thumb|The construction for the mid-point theorem's proof with similar triangles]]
'''Proposition:''' Let <math>D</math> and <math>E</math> be the midpoints of <math>AC</math> and <math>BC</math> in the triangle <math>ABC</math>. It is to be proved that,
# <math>DE \parallel AB</math> and;
# <math>DE = \frac {1}{2}AB</math>.
'''Construction:''' Add <math>D</math> and <math>E</math>, extend <math>DE</math> to <math>F</math> as <math>EF = DE</math>, and add <math>B</math> and <math>F</math>.

'''Proof:''' '''''[1]''''' In the triangles <math>\Delta CDE</math> and <math>\Delta EBF,</math>

<math>CE = BE</math> ; [Given]

<math>DE = EF</math> ; [According to the construction]

<math>\angle CED = \angle AEF</math> ; [Vertical Angles]

&there4; <math>\Delta CDE \cong \Delta EBF</math> ; [Side-Angle-Side theorem]

So, <math>\angle CDE = \angle BFE</math>

&there4; <math>CD \parallel BF</math>

Or, <math>AD \parallel BF</math> and <math>CD = BF = DA</math>

Therefore, <math>ADFB</math> is a parallelogram.

&there4; <math>DF \parallel AB</math> or <math>DE \parallel AB</math>


'''''[2]''''' <math>DF = AB</math>

Or <math>DE + EF = AB</math>

Or, <math>DE + DE = AB</math> [As, <math>\Delta CDE \cong \Delta EBF</math>]

Or, <math>2DE = AB</math>

Or, <math>DE = \frac {1}{2}AB</math>

&there4; In the triangle <math>\Delta ABC,</math> <math> DE \parallel AB</math> and <math>DE = \frac {1}{2}AB</math>, where <math>D</math> and <math>E</math> are the midpoints of <math>AC</math> and <math>BC</math>. '''''[Proved]'''''

=== Proof with the help of Coordinate Geometry ===
'''Proposition:''' Let <math>D</math> and <math>E</math> be the midpoints of <math>AC</math> and <math>AB</math> in the triangle <math>ABC</math>, where the coordinates of <math>A, B, C</math> are <math>A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)</math>. It is to be proved that,
# <math>DE = \frac {1}{2}BC</math> and
# <math>DE \parallel BC</math>
'''Proof:''' '''''[1]'''''
The distance of the segment <math>BC = \sqrt{(x_3 - x_2)^2 + (y_3 - y_2)^2}</math>

The midpoint of <math>A(x_1, y_1)</math> and <math>C(x_3, y_3)</math> is <math>D(\frac {x_1 + x_3}{2}, \frac {y_1 + y_3}{2})</math>.

In the same way, The midpoint of <math>A(x_1, y_1)</math> and <math>B(x_2, y_2)</math> is <math>E(\frac {x_1 + x_2}{2}, \frac {y_1 + y_2}{2})</math>

&there4; The distance of <math>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}</math>

<math>= \sqrt{(\frac {x_1 + x_3 - x_1 - x_2}{2})^2 + (\frac {y_1 + y_3 - y_1 - y_2}{2})^2}</math>

<math>= \sqrt{\frac {(x_3 - x_2)^2}{4} + \frac {(y_3 - y_2)^2}{4}}</math>

<math>= \sqrt{\frac {(x_3 - x_2)^2 + (y_3 - y_2)^2}{4}}</math>

<math>= \frac{\sqrt {(x_3 - x_2)^2 + (y_3 - y_2)^2}}{2}</math>

<math>= \frac {1}{2} BC</math> ; [As, <math>BC = \sqrt{(x_3 - x_2)^2 + (y_3 - y_2)^2}</math>]


'''''[2]'''''
The slope of <math>BC,</math>
<math>m_1 = \frac {y_2 - y_3}{x_2 - x_3}</math>

The slope of <math>DE,</math>
<math>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}}</math>
<math>= \frac {\frac {y_1 + y_2 - y_1 - y_3}{2}}{\frac {x_1 + x_2 - x_1 - x_3}{2}}</math>
<math>= \frac {\frac {y_2 - y_3}{2}}{\frac {x_2 - x_3}{2}}</math>
<math>= \frac {y_2 - y_3}{x_2 - x_3}</math>
<math>= m_1</math> ;
[As, <math>m_1 = \frac {y_2 - y_3}{x_2 - x_3}</math>]

Therefore, <math>DE \parallel BC</math>

&there4; In the triangle <math>\Delta ABC,</math> <math> DE \parallel BC</math> and <math>DE = \frac {1}{2}BC</math>, where <math>D</math> and <math>E</math> are the midpoints of <math>AC</math> and <math>AB</math>. '''''[Proved]'''''