Rhombus and circles angle problem

Question

Let $ABCD$ be a rhombus. The circle $(C_1)$ of center $B$ passing through $C$ and the circle $(C_2)$ of center $C$ passing through $B$. $E$ is one of the two points of $(C_1) \cap (C_2)$. The line $(ED)$ meets $(C_1)$ again in $F$. It is asked to find the measure of angle $\angle AFB$.

I tried a lot of angle chasing but in vain.

I even took a square instead just to find out a means to a solution but failed to get the value.

From geogebra, the measure would be $60^{\circ}$.

Answer 1

A fast approach using Euclidean geometry

1. Since $\mathcal C_1$ and $\mathcal C_2$ have the same radius, $\triangle BEC$ is equilateral.

2. $BE$ subtends on $\mathcal C_2$ the angle $\angle BDE$, which therefore has measure $30^\circ$, and \begin{eqnarray}\angle ADE &=& \frac12 \angle ADC + 30^\circ\\ &=& \frac12 \angle ABC + 30^\circ.\end{eqnarray}

3. On the other hand, $AE$ subtends on $\mathcal C_1$ the angle $\angle AFE$, by which we have \begin{eqnarray}180^\circ- \angle AFE &=& \frac12 \angle EBA=\\ &=& \frac12 \left(\angle ABC + 60^\circ\right).\end{eqnarray}
4. By 2. and 3. $\triangle AFD$ is isosceles, with $AD \cong AF$ and the thesis follows from the fact that $AB \cong AD \cong BF$.

EDIT (thanks to bjorn93 for his useful comment.) The situation when $ED<EF$ is shown below. However, as you can notice, the relationships found above do not change.

Answer 2

It should be clear that $\triangle BEC$ is equilateral. Denote $\angle BFD=\alpha$ and $\angle AFD=\beta$. We need to find $\alpha+\beta$. You can show that $\angle ADF=60^{\circ}-\alpha$ through some angle chasing. Sine law for $\triangle ADF$ gives you $$\frac{AD}{AF}=\frac{\sin\beta}{\sin(60^{\circ}-\alpha)} $$ and the sine law for $\triangle ABF$ gives you $$\frac{AB}{AF}=\frac{\sin(\alpha+\beta)}{\sin(2(\alpha+\beta))}=\frac{1}{2\cos(\alpha+\beta)} $$ Since $AB=AD$, $$\frac{\sin\beta}{\sin(60^{\circ}-\alpha)}=\frac{1}{2\cos(\alpha+\beta)} $$ Can you prove that the last equality implies $\beta=30^{\circ}$ or $\alpha+\beta=60^{\circ}$ and that $\beta=30^{\circ}$ is a contradiction?