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]