Napoleon’s Triangle

If we assume that one arrangement is possible as shown in the left figure, then it is possible for infinitely many arrangement. This can be proved as shown below :

Taking the left figure and name various points as shown in right figure. If we rotate those dotted lines about intersection point $\,\bf G$ then we get another arrangement of points $D,\,E,\,F$ . Also $\angle DGF = \angle FGE = \angle EGD = 120\degree$ . Since $\angle DGF = 120\degree$ and $\angle DBF = 60\degree$ we get that quadrilateral $DBFG$ is cyclic. Similarly other $2$ quadrilaterals namely $CFGE$ and $ADGE$ are also cyclic. All these three quadrilateral have only $1$ vertex in common namely $G$ . Hence all three circles will intersect at single point $G$ . Also if we move the intersection point G then we get different intersection point of circles.

In other words, if we take any point $G$ inside the triangle and make three lines $GD,\,GE,\,GF$ in the plane of triangle such that $\angle DGF = \angle FGE = \angle EGD = 120\degree$ to intersect the sides at $D,\,E,\,F$ then construction of such circles will always be possible. Hence taking different point $G$ and different orientation of lines $GD,\,GE,\,GF$ we get different points $D,\,E,\,F$ .

If $AD = BF = CE$ , then $m \angle DGF = m \angle FGE = m \angle EGD = 120°$ .

Proof: Let $T$ be the center of the equilateral triangle, and let $O$ be the center of the circle passing through $E$ , $G$ , $F$ , and $C$ . Also let $AD = BF = CE$ .

By symmetry, $\angle CFT = \angle AEG$ . By the properties of a straight line, $\angle CET = 180° - \angle AEG = 180° - \angle CFT$ . By the angle sum property of quadrilateral $CFTE$ , $\angle FTE + \angle CET + \angle ECF + \angle CFT = 360°$ , or $\angle FTE + (180° - \angle CFT) + 60° + \angle CFT = 360°$ , which solves to $\angle FTE = 120°$ .

Also by symmetry, $TE = TF$ . That means $\triangle FTE$ is an isosceles triangle, and since $\angle FTE = 120°$ , its base angles are $\angle TEF = \angle TFE = 30°$ .

Since central $\angle EOF$ subtends the same arc as inscribed $\angle ECF$ , $\angle EOF = 2 \cdot \angle ECF = 2 \cdot 60° = 120°$ . Since $\triangle EOF$ is an isosceles triangle, and since $\angle EOF = 120°$ , its base angles are $\angle OEF = \angle OFE = 30°$ .

Then $\triangle FTE \cong \triangle EOF$ by ASA congruence, which means $TE = TF = OE = OF$ . Since $TE = OE$ and $\angle TEO = \angle TEF + \angle OEF = 30° + 30° = 60°$ , $\triangle OTE$ is an equilateral triangle, and $OT = OE = OF$ . Therefore, $T$ is on circle $O$ .

By similar arguments, $T$ is on the other two circles as well. Therefore, $G$ and $T$ are the same point such that $m \angle DGF = m \angle FGE = m \angle EGD = 120°$ .

Napoleon's Theorem ( le petit empereur had no part in working this out) states that if you construct equilateral triangles on the sides of any triangle $DEF$ , the centres of these triangles form an equilateral triangle (in red). If we construct the circumcircles of these triangles, then they all meet at the point $G$ , where $\angle DGE = \angle EGF = \angle FGD = 120^\circ$ . The point $G$ is called the first Fermat point of $DEF$ . Since (for example) $\angle EAF = 60^\circ$ for any point $A$ on the major arc $EF$ , we can choose any point $A$ on this arc, and draw straight lines from $A$ through $E$ and $F$ to intersect the other circumcircles at $C$ and $B$ respectively. Then $ABC$ is an equilateral triangle with the desired properties. This works provided that all of the angles in $DEF$ are less than $120^\circ$ , which means that $G$ lies inside the triangle $DEF$ .