Quirky Convex Quadrilateral

Reflect $A$ with respect to $BD$ to get point $E$ . Since $\angle ADB = 30^\circ$ , hence $\angle ADE = 60^\circ$ and $AD=AE$ , so $ADE$ is equilateral. By side-side-side, It follows that $ABE$ is congruent to $BCD$ , hence $\angle BCD = \angle ABE = 2 \angle ABD = 150^\circ$ .

Since ABD is isosceles, $\angle ABD = 75^{\circ}, \angle ADB = 30^{\circ}$ . And let $\angle CBD = \angle CDB = a$ . Then $\angle BCD = 180 - 2a$ . By cosine rule,

$AB^{2} = AD^{2} + DB^{2} - 2AD \cdot DB \cos(30) = 2BD^{2}(1 - \frac{\sqrt{3}}{2})$ .

$BD^{2} = BC^{2} + CD^{2} - 2BC \cdot CD \cos(180 - 2a) = 2AB^{2}(1 - cos(180 - 2a)))$ .

Combining these two equations and simplifying gives $cos(180 - 2a) = -\frac{\sqrt{3}}{2}$ , so $\angle BCD = 150^{\circ}$ .

We have $\angle ADB=30^o$ . Now suppose that $AD=DB=x$ . Then $AB^2=2x^2-2x^2.\cos30^o$ $AB^2=(2-\sqrt{3})x^2$ So we have $DB^2=BC^2+DC^2-2BC.DC.\cos(\angle BCD)$ Since $BC^2=DC^2=BC.DC=AB^2=(2-\sqrt{3})x^2$ and $DB^2=x^2$ , we have $x^2=(2-\sqrt{3})x^2+(2-\sqrt{3})x^2-2.(2-\sqrt{3})x^2.\cos(\angle BCD)$ Eliminate $x^2$ we have $1=(4-2\sqrt{3})(1-\cos(\angle BCD))$ $\cos(\angle BCD)=1-\dfrac{1}{4-2\sqrt{3}}=\dfrac{-\sqrt{3}}{2}$ Since ABCD is a convex quadrilateral, then we have $\angle BCD = 150^o$

Let $\angle BCD = x,\, AB = a,$ and $AD =b$ . Since the triangle $ABD$ is isosceles, it follows that $\angle ABD = 75^\circ$ and $\angle ADB = 30^\circ.$ The law of cosines applied to the triangle $ABD$ and triangle $BCD$ yield $a^2 = b^2 + b^2 -2b^2\cos 30^\circ = b^2(2-\sqrt{3})$ and $b^2 = a^2+a^2-2a^2\cos x = 2a^2(1-\cos x),$ respectively. These two equations together imply that $\cos x =1-\frac{b^2}{2a^2} = 1-\frac{1}{2(2-\sqrt{3})}=-\frac{\sqrt{3}}{2}=\cos 150^\circ.$ Hence $x = \boxed{150^\circ}.$

Let AB=BC=CD=a and AD=DB=b

Apply Cosine Rule on Angle BAD and Angle ADB to get: (a/2b) = (1.732 - 1) / (2 1.414) and ((2 b b - a a) / (2 b b))=(1.732/2). Solve these two equations to get values of a and b and then apply Cosine Rule to Angle BCD to get cos(angle BCD) = -(1.732/2).

Therefore Angle BCD = 150 degrees.

Let BD = 1, by the cosine law in ABD, we can solve for $AB^2 = 2 - \sqrt{3}$ .

Then by the cosine law in BCD, we can find that $\angle BCD = 150^\circ$ .

By Law of Sines on triangle $ABD$, we have $$ \frac{AD}{\sin 75^\circ}=\frac{BA}{\sin 30^\circ}$$ So we can write $$ AD=(\sqrt{6}+\sqrt{2})k=BD, AB=2k=BC=CD. $$ Now by Law of Cosines on triangle $BCD$, we have $$ \cos C=\frac{(2k)^2+(2k)^2-[(\sqrt6+\sqrt2)k]^2}{2(2k)(2k)}$$ $$ =-\frac{\sqrt{3}}{2}\implies C=\boxed{120^\circ} $$

Since $\triangle ADB$ is isosceles, $m\angle ADB=30^\circ$ . By the Law of Sines, $\dfrac{\sin 30^\circ}{1}=\dfrac{\sin 75^\circ}{AD}$ Therefore, $AD=BD=\dfrac{\sqrt{6}+\sqrt{2}}{2}$ . Let $E$ be the foot of the perpendicular from $C$ to $\overline{BD}$ , and so $BE=ED$ , since $\triangle BCD$ is isosceles. $DE=\dfrac{\sqrt{6}+\sqrt{2}}{4}$ , and $m\angle ECD=\sin^{-1}\dfrac{\sqrt{6}+\sqrt{2}}{4}=75^\circ$ . Therefore, the answer is $\boxed{150}$ .

2(angle BAD)= angle BCD angle BAD=75 degree therefore angle BCD = 150 degree