# Timeless Theorems of Mathematics/Napoleon's theorem

!$\Delta LMN$ is an equilateral
triangle
The Napoleon\'s theorem states that if equilateral triangles are
constructed on the sides of a triangle, either all outward or all
inward, the lines connecting the centers of those equilateral triangles
themselves form an equilateral triangle. That means, for a triangle
$\Delta ABC$, if three equilateral triangles are constructed on the
sides of the triangle, such as $\Delta ACM$, $\Delta BCX$ and
$\Delta ABZ$ either all outward or all inward, the three lines
connecting the centers of the three triangles, $ML$, $LN$ and $MN$
construct an equilateral triangle $LMN$.

## Proof

!A trigonomatric proof of the Napoleon\'s
theorem.
Let, $\Delta ABC$ a triangle. Here, three equilateral triangles are
constructed, $\Delta BCE$, $\Delta ABD$ and $\Delta ACF$ and the
centroids of the triangles are $P$, $Q$ and $R$ respectively. Here,
$AC=b$, $AB=c$, $BC=a$, $PQ=r$, $PR=q$ and $QR=p$. Therefore, the area
of the triangle $\Delta ABC$, $T=\frac{1}{2}bc\cdot\sin A$
$\Rightarrow bc\cdot\sin A=2T$

For our proof, we will be working with one equilateral triangle, as
three of the triangles are similar (equilateral). A median of
$\Delta ACF$ is $AG = (m+k)$, where $AR = m$ and $RG = k$. $AQ=n$ and,
as $\Delta ACF$ is a equilateral triangle, $\angle AGC=90^\circ$.

Here, $m+k=\frac{b\sqrt{3}}{2}$. As the centroid of a triangle divides a
median of the triangle as $1:2$ ratio, then
$m=\frac{2}{3}\cdot\frac{b\sqrt{3}}{2}$ $=\frac{b}{\sqrt{3}}$.
Similarly, $n=\frac{c}{\sqrt{3}}$.

According to the Law of Cosines, $a^2 = b^2 + c^2 - 2bc\cdot \cos A$
(for $\Delta ABC$) and for $\Delta AQR$,
$p^2 = m^2 + n^2 - 2mn\cdot\cos(A+60^\circ)$

$= (\frac{b}{\sqrt{3}})^2 + (\frac{c}{\sqrt{3}})^2 - 2\frac{b}{\sqrt{3}}\cdot\frac{c}{\sqrt{3}}\cdot\cos(A+60^\circ)$

$= \frac{b^2 + c^2 - 2bc\cdot\cos(A+60^\circ)}{3}$

$= \frac{b^2 + c^2 - 2bc(\cos A\cdot cos60^\circ - \sin A\cdot\sin 60^\circ)}{3}$

$= \frac{b^2 + c^2 - 2bc(\cos A\cdot \frac{1}{2} - \sin A\cdot\frac{\sqrt{3}}{2})}{3}$

$= \frac{b^2 + c^2 - bc(\cos A\cdot - \sin A\cdot\sqrt{3})}{3}$

$= \frac{b^2 + c^2 - bc\cos A\cdot - 2T\sqrt{3})}{3}$

$= \frac{a^2 + 2bc\cos A - bc\cos A - 2T\sqrt{3})}{3}$ \[According to
the law of cosines for $\Delta ABC$\]

$= \frac{a^2 + bc\cos A - 2T\sqrt{3})}{3}$

$= \frac{a^2 + \frac {b^2 + c^2 - a^2}{2} - 2T\sqrt{3})}{3}$

$= \frac{\frac {b^2 + c^2 + a^2 - 4T\sqrt{3}}{2}}{3}$

Therefore, $p = \sqrt{\frac{1}{6} (a^2 + b^2 + c^2 - 4T\sqrt{3})}$

In the same way, we can prove,
$q = \sqrt{\frac{1}{6} (a^2 + b^2 + c^2 - 4T\sqrt{3})}$ and
$r = \sqrt{\frac{1}{6} (a^2 + b^2 + c^2 - 4T\sqrt{3})}$. Thus,
$p = q = r$.

$\therefore\Delta PQR$ is an equilateral triangle. ***\[Proved\]***

