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In the following figure $AF=BD=DC$ and $AE=EF$. Find the angle $\alpha$.

enter image description here

I tried so many different things (constructing triangles, drawing parallel lines, using Ceva theorem in triangles $ABD$ and $CBE$, ...). Any help for solving this will be much appreciated.

A geomtrical solution (without trigonometry) would be very nice.

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    $\begingroup$ @cosmo5 is it plausible, in your mind, that there is no typo? $\endgroup$ Mar 14, 2021 at 15:30
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    $\begingroup$ Its not a typo. $\endgroup$
    – Ghartal
    Mar 14, 2021 at 15:44
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    $\begingroup$ @cosmo5 The answer is exactly 60 degrees according to Geogebra. $\endgroup$
    – dodoturkoz
    Mar 14, 2021 at 15:46
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    $\begingroup$ @MathLover I made same mistake as you. It is AF=BD=DC not AD... $\endgroup$
    – cosmo5
    Mar 14, 2021 at 15:50
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    $\begingroup$ @Ghartal No, for $AF=BD=DC$ $\endgroup$
    – dodoturkoz
    Mar 14, 2021 at 17:09

2 Answers 2

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enter image description here

In isosceles $\triangle AEF$, drop $EG \perp AF$ with $G$ on $AF$. Let $AG=GF=x$. So $BD=2x$. Let $\angle BAD = \theta$. So $AE=x\sec \theta$.

Applying Menelaus' for transversal $CE$ to $\triangle ABD$, $$\frac{BC}{DC}\cdot\frac{DF}{FA}\cdot\frac{AE}{EB}=1$$ $$\Rightarrow DF=EB\cos \theta$$ $$\Rightarrow AD-2x=(AB-x\sec\theta)\cos \theta$$ $$\Rightarrow AD=AB\cos \theta+x$$

Next drop $BH \perp AD$ with $H$ on $AD$. In right $ABH$, $AH=AB \cos \theta$. Then $AD=AH+DH$ implies $DH=x$!

Hence in right triangle $BHD$, $BD=2x$, $HD=x=BD/2$. It turns out $\triangle BHD$ is $30^{\circ}-60^{\circ}-90^{\circ}$ and we conclude $$\alpha = 60^{\circ}$$

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  • $\begingroup$ Cool trigonometric proof. There is also a geometric proof, if you are curios. $\endgroup$ Mar 20, 2021 at 19:02
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Here is a purely geometric proof without any trigonometry.

enter image description here

Draw the three lines $a, \, b, \, c$ such that: $$A \, \in \, a \,\,\text{ and } \,\, a \, || \, BC$$ $$B \, \in \, b \,\,\text{ and } \,\, b \, || \, CA$$ $$C \, \in \, c \,\,\text{ and } \,\, c \, || \, AB$$ Furthermore, let $$B^* \, = \, a \, \cap \, c \,\,\, \text{ and } \,\,\, A^* = b \, \cap \, c$$ Then, $ABA^*C$ and $ABCB^*$ are parallelograms and furthermore $$AB = CA^* = CB^*$$ as well as $$AB^* = BC$$ Apply Menelaus' theorem to the triangle $BCE$ intersected by the line that passes through the collinear points $A, \, F,\, D$: $$\frac{AB}{AE} \cdot \frac{EF}{CF} \cdot \frac{DC}{BD} = 1$$ Since by assumption $AE = EF$ and $BD = CD$, one concludes that $$AB = CF$$ Combining this latter fact with the earlier conclusions, one observes that $$CF = AB = CA^* = CB^*$$ which is possible if and only if triangle $\Delta \, A^*B^*F$ has $90^{\circ}$ angle at vertex $F$, i.e. $\angle \, A^*FB^* = 90^{\circ}$. But that means that: $$\angle \, AFB^* = 180^{\circ} - \angle \, A^*FB^* = 180^{\circ} - 90^{\circ} = 90^{\circ}$$ Recall that by assumption $AF = BD = DC = \frac{1}{2}\,BC$. But by construction, $ABCB^*$ is a parallelogram and $AB^* = BC$ so $$\Delta \, AFB^* \, \text{ is a triangle with the properties that } \, \angle \, AFB^* = 90^{\circ} \,\, \text{ and } \,\, AF = \frac{1}{2}\, AB^* $$
which is possible if and only if $$\angle \, FAB^* = 60^{\circ}$$ However, $AB^*$ is parallel to $BC$ and consequently, by the properties of a pair of parallel lines intersected by a third line, $$\angle \, ADB = \angle \, DAB^* = \angle \, FAB^* = 60^{\circ}$$

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  • $\begingroup$ You are right, I got it now! I neglected the collinear requirement and you did give a beautiful answer. I will upvote yours and delete my answer. Thank you for the challenge. $\endgroup$ Mar 20, 2021 at 19:21
  • $\begingroup$ Nice one. You also find a $30-60-90$ triangle using construction. $\endgroup$
    – cosmo5
    Mar 21, 2021 at 5:18

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