Ratios!

$A_{\triangle ABC} = A_{\triangle DBC} + A_{\triangle ADC} = \frac{1}{2}az \sin \lambda + \frac{1}{2}bz \sin \lambda = \frac{1}{2}z(a + b)\sin \lambda$ , so $\cfrac{z(a + b)\sin \lambda}{A_{\triangle ABC}} = \cfrac{z(a + b)\sin \lambda}{\frac{1}{2}z(a + b)\sin \lambda} = \boxed{2}$ .

$m\angle{CDB} = 180^{\circ} - (B + \lambda)$

Applying the law of sines on $\triangle{ABC} \implies$

$\dfrac{b}{\sin(B)} = \dfrac{c}{\sin(2\lambda)} \implies \sin(B) = \dfrac{b\sin(2\lambda)}{c}$

Appling the law of cosines on $\triangle{ABC} \implies c^2 = a^2 + b^2 - 2ab\cos(2\lambda) \implies$

$\sin(B) = \dfrac{b\sin(2\lambda)}{\sqrt{a^2 + b^2 - 2ab\cos(2\lambda)}} \implies$ $\cos(B) = \dfrac{a - b\cos(2\lambda)}{\sqrt{a^2 + b^2 - 2ab\cos(2\lambda)}}$

and applying the law of sines on $\triangle{BCD} \implies \dfrac{z}{\sin(B)} = \dfrac{a}{\sin(B + \lambda)} \implies$

$z = \dfrac{a\sin(B)}{\sin(B)\cos(\lambda) + \cos(B)\sin(\lambda)}$

Using the above

$\sin(B) = \dfrac{b\sin(2\lambda)}{\sqrt{a^2 + b^2 - 2ab\cos(2\lambda)}}$

and

$\cos(B) = \dfrac{a - b\cos(2\lambda)}{\sqrt{a^2 + b^2 - 2ab\cos(2\lambda)}}$

and simplifying we obtain $\boxed{z = (\dfrac{2ab}{a + b})\cos(\lambda)}$

$h = a\sin(B) = \dfrac{ab\sin(2\lambda)}{\sqrt{a^2 + b^2 - 2ab\cos(2\lambda)}} \implies$ $\boxed{A_{\triangle{ABC}} = \dfrac{1}{2}ab\sin(2\lambda)} \implies$

$\dfrac{z(a + b)\sin(\lambda)}{A_{\triangle{ABC}}} = \boxed{2}$ .