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 $$\Delta ABC$$, 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
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]