CIMC 2015 Quadrangle

We note that:

$\begin{aligned} \angle ADC & = 180^\circ - 25^\circ - 25^\circ = 130^\circ \\ \Rightarrow \angle ABC & = 360^\circ - \angle BAD - \angle ADC - \angle DCB \\ & = 360^\circ - (85^\circ+25^\circ) - 130^\circ - (25^\circ + 30 ^\circ) = 65^\circ \end{aligned}$

Let $\angle DBC = \theta\quad \Rightarrow \angle ABD = 65^\circ - \theta$ .

Using Sine Rule, we have:

$\begin{aligned} \dfrac{\color{#3D99F6}{AD}}{BD} = \dfrac{\sin{(65^\circ - \theta)}}{\sin{110^\circ}} & = \dfrac{\color{#3D99F6}{CD}}{BD} = \dfrac{\sin{\theta}}{\sin{55^\circ}} \quad \quad \quad \small \color{#3D99F6} {[AD=CD]} \\ \Rightarrow \dfrac{\sin{(65^\circ - \theta)}}{2 \sin{55^\circ} \cos{55 ^\circ}} & = \dfrac{\sin{\theta}}{\sin{55^\circ}} \\ \sin{(65^\circ - \theta)} & = 2 \cos{55 ^\circ} \sin{\theta} \\ \sin{65^\circ} \cos{\theta} - \cos{65^\circ} \sin{\theta} & = 2 \cos{55 ^\circ} \sin{\theta} \\ (2 \cos{55 ^\circ} + \cos{65^\circ} ) \sin{\theta} & = \sin{65^\circ} \cos{\theta} \\ \Rightarrow \tan{\theta} & = \dfrac {\sin{65^\circ}} {2 \cos{55 ^\circ} + \cos{65^\circ}} \\ & = \dfrac {\sin{(60^\circ+5^\circ)}} {2 \cos{(60^\circ - 5^\circ)} + \cos{(60^\circ + 5^\circ)}} \\ & = \dfrac {\frac{\sqrt{3}}{2} \cos{5^\circ} + \frac{1}{2} \sin{5^\circ}} {2(\frac{1}{2} \cos{5^\circ} + \frac{\sqrt{3}}{2} \sin{5^\circ}) + \frac{1}{2} \cos{5^\circ} - \frac{\sqrt{3}}{2} \sin{5^\circ} } \\ & = \dfrac {\sqrt{3} \cos{5^\circ} + \sin{5^\circ}} {3\cos{5^\circ} + \sqrt{3} \sin{5^\circ}} \\ & = \dfrac{1}{\sqrt{3}} \\ \Rightarrow \theta & = 30^\circ \end{aligned}$

$$

$\begin{aligned} \Rightarrow \angle BDC & = 180^\circ - \angle DCB - \angle DBC \\ & = 180^\circ - (25^\circ + 30^\circ) - 30^\circ \\ & = \boxed{95^\circ} \end{aligned}$

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