Medians of Triangle Form Six Triangles of Equal Area
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Theorem
Let $\triangle ABC$ be an arbitrary triangle.
Draw $AL$, $BM$ and $CN$, the medians of $\triangle ABC$.
Then $\triangle ABC$ is divided into six small triangles:
- $\triangle GAN$
- $\triangle GBN$
- $\triangle GBL$
- $\triangle GCL$
- $\triangle GCM$
- $\triangle GAM$
Each one of these six triangles has the same area.
Proof
- $AN = BN$
- $BL = CL$
- $CM = AM$
Each pair of triangles with bases on the same side of $\triangle ABC$ consists of two triangles:
It follows by Triangles with Equal Base and Same Height have Equal Area:
- each member of the pair has the same area.
| \(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \AA \paren { GAN }\) | \(=\) | \(\ds \AA \paren { GBN }\) | ||||||||||
| \(\text {(2)}: \quad\) | \(\ds \AA \paren { GCM }\) | \(=\) | \(\ds \AA \paren { GAM }\) | mutatis mutandis | ||||||||||
| \(\text {(3)}: \quad\) | \(\ds \AA \paren { GBL }\) | \(=\) | \(\ds \AA \paren { GCL }\) | mutatis mutandis |
Now consider the pair $\triangle CAN$ and $\triangle CBN$:
- they have bases of equal magnitude
- they have the same vertex $C$
- so they have the same altitude and equal height.
By Triangles with Equal Base and Same Height have Equal Area:
- $\AA \paren { CAN } = \AA \paren { CBN }$
It follows that the component triangles of $\triangle CAN$ and $\triangle CBN$ add up to the same total area.
| \(\ds \leadsto \ \ \) | \(\ds \AA \paren { GAN } + \AA \paren { GAM } + \AA \paren { GCM }\) | \(=\) | \(\ds \AA \paren { GBN } + \AA \paren { GBL } + \AA \paren { GCL }\) | |||||||||||
| \(\ds \AA \paren { GAM } + \AA \paren { GCM }\) | \(=\) | \(\ds \AA \paren { GBL } + \AA \paren { GCL }\) | Common Notion $2$ with $(1)$ | |||||||||||
| \(\ds 2 \cdot \AA \paren { GCM }\) | \(=\) | \(\ds 2 \cdot \AA \paren { GCL }\) | Common Notion $1$ with $(2)$ and $(3)$ | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \AA \paren { GCM }\) | \(=\) | \(\ds \AA \paren { GCL }\) | |||||||||||
| \(\ds \AA \paren { GAM }\) | \(=\) | \(\ds \AA \paren { GAN }\) | mutatis mutandis | |||||||||||
| \(\ds \AA \paren { GBL }\) | \(=\) | \(\ds \AA \paren { GBN }\) | mutatis mutandis | |||||||||||
| \(\ds \AA \paren { GCM }\) | \(=\) | \(\ds \AA \paren { GCL }\) | mutatis mutandis |
The result follows.
$\blacksquare$