Prove, using affine geometry, that in this figure $\Delta DEF$ is always equilateral

Question

Consider the following figure:

$\Delta DAC, \Delta CEB, \Delta AFB$ are isosceles. $\angle ADC = \angle CEB = \angle AFB = 120^{\circ}$.

Prove that $\Delta DEF$ is equilateral.

Now, there is a proof for this using coordinate geometry. And its an ugly one. I was wondering if there was a 'better' proof for this using affine geometry.

We are required to prove that:

$$||\vec{DE}|| = ||\vec{EF}|| = ||\vec{FD}||$$

$$\implies (\vec{E} - \vec{D})\cdot (\vec{E} - \vec{D}) = (\vec{F} - \vec{E})\cdot (\vec{F} - \vec{E}) = (\vec{D} - \vec{F})\cdot (\vec{D} - \vec{F})$$

$$\implies 2\cdot\vec{E} = \vec{F} + \vec{D}$$

$$\implies 2\cdot\vec{D} = \vec{F} + \vec{E}$$

$$\implies 2\cdot\vec{F} = \vec{D} + \vec{E}$$

How should I proceed?

EDIT: Thanks for all the wonderful solutions below, but I require a solution through affine geometry only.

Answer 1

A hint: Show that the red and the blue triangles are congruent.

Answer 2

Hint: Draw equilateral triangles at $A$ and $B$:

Answer 3

This is how I look at the problem:

Since $\overline{AO_1}=\overline{CO_1}$, $\overline{AO_2}=\overline{BO_2},$ and $\overline{BO_3}=\overline{CO_3}$, we can draw circles with centers $O_1,\ O_2,$ and $O_3$ intersecting points $A,C$; $A,B$; and $B,C$, respectively. Now, the first thing we notice is that all three of these circles intersect at a point $D$. How do we prove that?

Let $D_1$ be the intersection of $O_1$ and $O_2$ that is not $A$, $D_2$ be that of $O_1$ and $O_3$ that is not $C$, and $D_3$ be that of $O_2$ and $O_3$ that is not $B$. Note that $\angle AO_1 D_1=2\angle ACD_1$. Meanwhile $\angle D_1O_2B=2\angle D_1AB=2\angle D_1 AC$. Note that $\angle D_1AC+\angle D_1CA+ \angle CD_1A=180^\circ$. We then have $\frac{1}{2}\angle D_1 O_2 B+\frac{1}{2}\angle D_1 O_1 A+ (360-2\angle AO_1C)=180^\circ$, which ultimately simplifies to $\angle D_1O_2B+\angle D_1O_1A=120^\circ$. Similarly, we obtain $\angle D_2O_1C+\angle D_2O_3B=\angle D_3O_2B+\angle D_3O_3B=120^\circ$.

Therefore $\angle D_1O_1A+\angle CO_1D_2+\angle BO_2D_1+\angle D_2O_3B=240^\circ$ or $\angle BO_2D_1+\angle D_2O_3B=120^\circ-\angle D_2O_1D_1$ which ultimately implies $\angle D_2O_1D_1+\angle D_1O_2D_3+\angle D_3O_3D_2=0$. Since $D_1, D_3,$ and $A$ lie on $O_2$ and $D_1, D_2$, and $A$ lie on $O_1$, $2\angle D_2AD_1+2\angle D_1AD_3+\angle D_3O_3 D_2=2\angle D_2AD_3+\angle D_3O_3D_2=0$. However, this implies $A$ lies on the circle with center $O_3$ unless $D_2=D_3$, therefore $D:=D_1=D_2=D_3$.

Now for the fun part. Note that $\overline{O_1O_2}$ is an angle bisector of $\angle DO_1A$ and $\angle AO_2D$ since the quadrilateral $AO_2DO_1$ forms a kite. Similarly $\overline{O_1O_3}$ bisects $\angle CO_1D$ and $\angle DO_3C$ and $\overline{O_2O_3}$ bisects $\angle BO_2D$ and $\angle DO_3B$. So $$\angle O_3O_1O_2=\angle O_3O_1D+\angle DO_1O_2=\frac{1}{2}(\angle DO_1A+\angle CO_1D)=\frac{1}{2}(120^\circ)=60^\circ$$ $$ \angle O_2O_3O_1=\angle DO_3O_1-\angle DO_3O_2=\frac{1}{2}(\angle DO_3C-\angle DO_3B)=\frac{1}{2}(120^\circ)=60^\circ $$ $$ \angle O_1O_2O_3=\angle O_1O_2D-\angle O_3O_2D=\frac{1}{2}(\angle AO_2D-\angle BO_2D)=\frac{1}{2}(120^\circ)=60^\circ $$

$\Rightarrow$ $\triangle O_1O_2O_3$ is equiangular

Note: Sorry for all the angle bash!