Timeless Theorems of Mathematics/Napoleon's theorem

$\mathrm{\Delta }LMN$ 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 $\mathrm{\Delta }ABC$, if three equilateral triangles are constructed on the sides of the triangle, such as $\mathrm{\Delta }ACM$, $\mathrm{\Delta }BCX$ and $\mathrm{\Delta }ABZ$ 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, $\mathrm{\Delta }ABC$ a triangle. Here, three equilateral triangles are constructed, $\mathrm{\Delta }BCE$, $\mathrm{\Delta }ABD$ and $\mathrm{\Delta }ACF$ 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 $\mathrm{\Delta }ABC$, $T=\frac{1}{2}bc\cdot \sin A$ $\Rightarrow bc\cdot \sin A=2T$

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

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 $\mathrm{\Delta }ABC$) and for $\mathrm{\Delta }AQR$, $p^2=m^2+n^2-2mn\cdot \cos (A+60^{\circ })$

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

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

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

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

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

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

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

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

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

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

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, $p=q=r$.

$\therefore \mathrm{\Delta }PQR$ is an equilateral triangle. [Proved]