South African Mathematics Olympiad

We have $\angle FAD=\angle BAD-\angle BAF=90^{\circ }-60^{\circ }=30^{\circ }$. Since $AF=AB=AD$, triangle $AFD$ is isosceles, which means that $\angle ADF=\angle AFD=\frac{180^{\circ }-\angle FAD}{2}=75^{\circ }$ and $\angle CDF=\angle CDA-\angle ADF=90^{\circ }-75^{\circ }=15^{\circ }$. By symmetry, we also have $\angle ADE=15^{\circ }$, thus $\angle EDF=90^{\circ }-\angle ADE-\angle CDF=60^{\circ }$.

Again by symmetry (with respect to the diagonal $BD$), $DE=DF$, so $DEF$ is an isosceles triangle with an angle of $60^{\circ }$. Therefore, $DEF$ is indeed an equilateral triangle.

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Alternative solution.

Let $G$ and $H$ be the midpoints of $AB$ and $CD$ respectively, and let $M$ be the centre of the square. We denote the side length of the square and the two equilateral triangles by $a$. By Pythagoras' Theorem, $$G{F}^2=A{F}^2-A{G}^2=a^2-{\left(\frac{a}{2}\right)}^2=\frac{3a^2}{4},$$ so $GF=\frac{\sqrt{3}a}{2}$. Next we find $FH=GH-GF=a-\frac{\sqrt{3}a}{2}=\frac{(2-\sqrt{3})a}{2}$, $FM=GF-GM=\frac{\sqrt{3}a}{2}-\frac{a}{2}=\frac{(\sqrt{3}-1)a}{2}$ and by symmetry $EM=FM=\frac{(\sqrt{3}-1)a}{2}$.

Applying Pythagoras' Theorem again, we obtain $$D{F}^2=D{H}^2+F{H}^2={\left(\frac{a}{2}\right)}^2+{\left(\frac{(2-\sqrt{3})a}{2}\right)}^2=a^2\left(\frac{1}{4}+\frac{4-4\sqrt{3}+3}{4}\right)=a^2(2-\sqrt{3})$$ and $$E{F}^2=E{M}^2+F{M}^2=2{\left(\frac{(\sqrt{3}-1)a}{2}\right)}^2=2a^2\cdot \frac{3-2\sqrt{3}+1}{4}=a^2(2-\sqrt{3}).$$ Thus $DF=EF$, and by symmetry $DE=EF$. This means that $DEF$ is an equilateral triangle.