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The triangle formed by the three parallel lines through the three midpoints of sides of a triangle is called its [[medial triangle]].
==Proof==
===Proof by construction===
{{Math proof|proof=[[File:Midpoint Theorem proof.png|thumb|304x304px]]
'''Given''': In a <math>\triangle ABC </math> the points M and N are the midpoints of the sides AB and AC respectively.
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Hence BCDM is a [[parallelogram]]. BC and DM are also equal and parallel.
*<math>MN\parallel BC</math>
*<math>MN={1\over 2}MD={1\over 2}BC</math>,
[[Q.E.D.]]
}}
===Proof by similar triangles===
{{Math proof|proof=[[File:Midpoint theorem.svg|thumb|304x304px]]
Let D and E be the midpoints of AC and BC.
'''To prove:'''
* <math>DE\parallel AB</math>,
* <math>DE = \frac{1}{2}AB</math>.
'''Proof:'''
<math>\angle C</math> is the common angle of <math>\triangle ABC</math> and <math>\triangle DEC</math>.
Since DE connects the midpoints of AC and BC, <math>AD=DC</math>, <math>BE=EC</math> and <math>\frac{AC}{DC}=\frac{BC}{EC}=2.</math> As such, <math>\triangle ABC</math> and <math>\triangle DEC</math> are [[Similarity (geometry)|similar]] by the SAS criterion.
Therefore, <math>\frac{AB}{DE}=\frac{AC}{DC}=\frac{BC}{EC}=2,</math> which means that <math>DE=\frac{1}{2}AB.</math>
Since <math>\frac{AB}{DE}=\frac{AC}{DC}=\frac{BC}{EC},</math> DE is parallel to AB by [[Intercept theorem]].
[[Q.E.D.]]
}}
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