Timeless Theorems of Mathematics/Napoleon's theorem

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 ${\displaystyle \Delta ABC}$, if three equilateral triangles are constructed on the sides of the triangle, such as ${\displaystyle \Delta ACM}$, ${\displaystyle \Delta BCX}$ and ${\displaystyle \Delta ABZ}$ either all outward or all inward, the three lines connecting the centers of the three triangles, ${\displaystyle ML}$, ${\displaystyle LN}$ and ${\displaystyle MN}$ construct an equilateral triangle ${\displaystyle LMN}$.

Let, ${\displaystyle \Delta ABC}$ a triangle. Here, three equilateral triangles are constructed, ${\displaystyle \Delta BCE}$, ${\displaystyle \Delta ABD}$ and ${\displaystyle \Delta ACF}$ and the centroids of the triangles are ${\displaystyle P}$, ${\displaystyle Q}$ and ${\displaystyle R}$ respectively. Here, ${\displaystyle AC=b}$, ${\displaystyle AB=c}$, ${\displaystyle BC=a}$, ${\displaystyle PQ=r}$, ${\displaystyle PR=q}$ and ${\displaystyle QR=p}$. Therefore, the area of the triangle ${\displaystyle \Delta ABC}$, ${\displaystyle T={\frac {1}{2}}bc\cdot \sin A}$ ${\displaystyle \Rightarrow bc\cdot \sin A=2T}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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