Where did that Algebra come from?

Let the width and height of rectangle $ABCD$ be $w$ and $h$ respectively, and $DL = x$ and $BK=y$ .

Since $[ABK]=[ADL]$ , $\implies \dfrac {wy}2 = \dfrac {hx}2$ , $\implies wy = hx$ , $\implies \dfrac yh = \dfrac xw$ , $\implies y= \dfrac {hx}w$ .

We note that:

$\begin{aligned} 3[ADL] + [CKL] & = [ABCD] \\ \frac {3hx}2 + \frac {(w-x)(h-y)}2 & = wh \\ 3hx + wh -hx-{\color{#3D99F6}wy}+xy & = 2wh & \small \color{#3D99F6} \text{Note that }wy=hx \\ hx + xy - wh & = 0 & \small \color{#3D99F6} \text{Dividing both sides by }wh \\ \frac xw + \frac xw\cdot{\color{#3D99F6}\frac yh} - 1 & = 0 & \small \color{#3D99F6} \text{Note that }\frac yh = \frac xw \\ \implies \left(\frac xw\right)^2 + \frac xw -1 & = 0 \\ \implies \frac xw & = \frac {\sqrt 5 -1}2 \end{aligned}$

Therefore, $\dfrac {DL}{DC} + \dfrac {BK}{BC} = \dfrac xw + \dfrac yh = 2 \cdot \dfrac xw = \sqrt 5 -1$ . $\implies a+b = \boxed{6}$