Adventitious Quadrangle 3

This solution will be attempted without trigonometry.

1. reflect $\Delta ABD$ about $AD$ .

2. show that $CAB'$ is a straight line.

3. show that $\Delta BDB'$ is an equilateral triangle.

4. show that $BCDB'$ is cyclic by inscribed angle theorem.

5. show that $x=60^{\circ}$ .

Use a 12 sided regular polyhedron to help with this one and do a bunch of angle chasing. The key to this proof is the symmetrical kite, the one with a right angle vertex.

Isn't drawing triangles trigonometry? Anyway, I know what @Albert Lau means. I am using trigonometry " proper " here.

Let $AC$ and $BD$ intersect at $E$ . We note that $\triangle ABC$ is a $30^\circ$ - $60^\circ$ - $90^\circ$ right triangle. Let $BC=1$ ; then $AC=2$ . Let $CE=a$ ; then by sine rule, we have $\dfrac a{\sin 45^\circ} = \dfrac 1{\sin \angle BEC} = \dfrac 1{\sin 75^\circ}$ , $\implies \color{#3D99F6} a = \dfrac {\sin 45^\circ}{\sin 75^\circ}$ .

Now, let $\angle CDB = \theta$ and $CD=b$ . By sine rule again,

$\begin{cases} \dfrac {\sin \angle CDB}{CE} = \dfrac {\sin \theta}a = \dfrac b{\sin \angle CED} = \dfrac b{\sin 105^\circ} = \dfrac b{\sin 75^\circ} \\ \dfrac {\sin \angle ADC}{AC} = \dfrac {\sin (\theta+30^\circ)}2 = \dfrac b{\sin \angle CAD} = \dfrac b{\sin 75^\circ} \end{cases}$

$\begin{aligned} \implies \frac {\sin \theta}{\color{#3D99F6}a} & = \frac {\sin (\theta + 30^\circ)}2 & \small \color{#3D99F6} \text{Note that } a = \frac {\sin 45^\circ}{\sin 75^\circ} \\ \frac {\sin \theta \sin 75^\circ}{\sin 45^\circ} & = \frac {\sin \theta \cos 30^\circ + \cos \theta \sin 30^\circ}2 \\ 2\sin 75^\circ \sin \theta & = \sin 45^\circ \cos 30^\circ \sin \theta + \sin 45^\circ \sin 30^\circ \cos \theta \\ \implies \tan \theta & = \frac {\sin 45^\circ \sin 30^\circ}{2\sin 75^\circ - \sin 45^\circ \cos 30^\circ} \end{aligned}$