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[parent] triangle mid-segment theorem (Theorem)

Theorem. The segment connecting the midpoints of any two sides of a triangle is parallel to the third side and is half as long.


\begin{pspicture}(-1,-0.5)(5,5.5) \psline(0,0)(4,0) \psline[arrows=->,arrowsize=... ...5,2.5) \psline[arrows=->,arrowsize=5pt,linecolor=blue](0,0)(3,5) \end{pspicture}

Proof. In the triangle $ ABC$, let $ A'$ be the midpoint of $ AC$ and $ B'$ the midpoint of $ BC$. Using the side-vectors $ \overrightarrow{AC}$ and $ \overrightarrow{CB}$ as a basis of the plane, we calculate the mid-segment $ A'B'$ as a vector:

$\displaystyle \overrightarrow{A'B'} \,=\, \overrightarrow{A'C}+\overrightarrow{... ...}(\overrightarrow{AC}+\overrightarrow{CB}) \,=\, \frac{1}{2}\overrightarrow{AB}$
The last expression indicates that the segment $ A'B'$ is such as asserted.

Corollary. If one connects the midpoints of the adjacent sides of a quadrilateral, one obtains a parallelogram.



"triangle mid-segment theorem" is owned by pahio.
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See Also: mutual positions of vectors, parallelogram theorems, median of trapezoid, common point of triangle medians

Other names:  mid-segment theorem

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Cross-references: parallelogram, quadrilateral, expression, vector, calculate, plane, parallel, triangle, sides, midpoints, segment
There are 2 references to this entry.

This is version 5 of triangle mid-segment theorem, born on 2008-02-05, modified 2008-05-19.
Object id is 10234, canonical name is TriangleMidSegmentTheorem.
Accessed 1015 times total.

Classification:
AMS MSC: 51M04 (Geometry :: Real and complex geometry :: Elementary problems in Euclidean geometries)
 51M25 (Geometry :: Real and complex geometry :: Length, area and volume)
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