A Golden Problem!

Let $AB=DC=1$ and $AC=x$ . By angle bisector theorem ,

$\dfrac {BD}{AB} = \dfrac {DC}{AC} \implies \dfrac {BD}1 = \dfrac 1x \implies BC = 1 + BD = 1 + \dfrac 1x$

By sine rule ,

$\begin{aligned} \frac {\sin B}{\sin C} & = \frac {AC}{AB} = x \\ \frac {\sin (2C)}{\sin C} & = x \\ 2\sin C \cos C & = x \sin C \\ \implies \cos C & = \frac x2 \end{aligned}$

By cosine rule ,

$\begin{aligned} AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos C & = AB^2 \\ x^2 + \left(1+\frac 1x \right)^2 - 2 x \left(1+\frac 1x \right) \cdot \frac x2 & = 1 \\ x^2 + 1 + \frac 2x + \frac 1{x^2} - x^2 - x & = 1 \\ \frac 2x + \frac 1{x^2} - x & = 0 \\ x^3 & = 2x + 1 \\ \implies x & = \varphi & \small \blue{\text{where }\varphi = \frac {1+\sqrt 5}2 \text{ denotes the golden ratio,}} \end{aligned}$

Then $\cos C = \dfrac \varphi 2 \implies C = 36^\circ$ . And the area of $\triangle ABC$ is given by:

$\begin{aligned} A_\triangle & = \frac 12 \cdot AC \cdot BC \cdot \sin C = \frac \varphi 2 \left(1 + \frac 1\varphi \right) \sqrt{1-\frac {\varphi^2}4} \\ \frac {A_\triangle}{BC^2} & = \frac {\varphi+1}{2\varphi} \sqrt{1-\frac {\varphi^2}4} = \frac {\varphi^2\sqrt{4-\varphi^2}}{4\varphi^2} = \frac 14 \sqrt{4-\frac {3+\sqrt 5}2} \\ & = \frac 14 \sqrt{\frac {5-\sqrt 5}2} = \frac 14 \sqrt{\sqrt 5 \left(\frac {\sqrt 5-1}2\right)} \\ & = \frac 14 \sqrt{\sqrt 5 \left(\frac {\sqrt 5+1}2 - 1\right)} = \frac {\sqrt{\sqrt 5(\varphi -1)}}4 \end{aligned}$

Therefore $a+b = 5 + 4 = \boxed 9$ .

$\theta + \lambda + 180^{\circ} - m = 180^{\circ} \implies m = \theta + \lambda$

Using law of sines of $\triangle{ADC}$ and $\triangle{ABD} \implies$

$\dfrac{\overline{DC}}{\sin(\theta)} = \dfrac{\overline{AD}}{\sin(\lambda)}$

and

$\dfrac{\overline{AB}}{\sin(\theta + \lambda)} = \dfrac{\overline{AD}}{\sin(2\lambda)}$

$\implies \dfrac{\overline{AB}\sin(2\lambda)}{\sin(\theta + \lambda)} =$ $\dfrac{\overline{DC}\sin(\lambda)}{\sin(\theta)}$

$\implies \sin(2\lambda)\sin(\theta) = \sin(\lambda)\sin(\theta + \lambda) \implies$

$2\cos(\lambda)\sin(\theta) = \sin(\theta)\cos(\lambda) + \cos(\theta)\sin(\lambda)$

$\implies \cos(\theta)\sin(\lambda) - \cos(\lambda)\sin(\theta) = 0 \implies \sin(\lambda - \theta) = 0 \implies \theta = \lambda$

$\implies 180^{\circ} = 3\lambda + 2\theta = 5\lambda \implies \lambda = 36^{\circ} = \theta \implies 2\theta = 2\lambda = 72^{\circ}$

$h = \overline{AC}\sin(36^{\circ}) = \overline{BC}\sin(36^{\circ}) \implies$

$A_{\triangle{ABC}} = \dfrac{1}{2}(\overline{BC})^2\sin(36^{\circ})$ $\implies \dfrac{A_{\triangle{ABC}}}{(\overline{BC})^2} = \dfrac{1}{2}\dfrac{\sqrt{10 - 2\sqrt{5}}}{4} =$

$= \dfrac{\sqrt{2(5 - \sqrt{5})}}{8} = \dfrac{\sqrt{4\sqrt{5}(\dfrac{\sqrt{5} - 1}{2})}}{8}$ $= \dfrac{\sqrt{\sqrt{5}(\phi - 1)}}{4} =$

$\dfrac{\sqrt{\sqrt{a}(\phi - 1)}}{b} \implies a + b = \boxed{9}$ .