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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 $60^{\circ}$.enter image description here

ahmed
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2 Answers2

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A fast approach using Euclidean geometry

enter image description here

  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.

enter image description here

dfnu
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    @ahmed Note that there's actually two cases depending on the relative position of D, E, and F. In the second case, $\angle ADE= \frac 12\angle ABC-30^{\circ}$. – bjorn93 Dec 19 '19 at 21:37
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    @bjorn93 edited; check it out. I think the relationship between $\angle ADE$ and $\angle ABC$ does not change. – dfnu Dec 19 '19 at 21:53
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    well, there's actually more than 2 cases for the position of the points. For example, D and E can be in the same half-plane with respect to BC and $\angle BDE$ can be $150^{\circ}$. What I meant is that they give you 2 cases for the angles: $\angle ADE=\frac 12\angle ABC\pm 30^{\circ}$. – bjorn93 Dec 19 '19 at 22:28
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    @bjorn93 ah ok! Got it. Now I turned off my PC, but as soon as possible I'll show one of the cases you mention. Thanks again! – dfnu Dec 19 '19 at 22:31
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    I don't think it's necessary. In any case, we get the same answer and the solution is similar to yours with slight adjustments. – bjorn93 Dec 19 '19 at 22:32
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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?

bjorn93
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