Algrabric manipulation

To satisfy the first equation and the quantities being positive, let $a=r \cos \theta$ , $b=r \sin \theta$ , $c=r \cos \phi$ and $d=r \sin \phi$ , where $r>0$ and $\theta$ and $\phi$ are in the interval $\left(0,\frac{\pi}{2}\right)$ .

The second equation then becomes $r^2 \cos^2 \theta + r^2 \sin^2 \phi - r^2 \cos \theta \cdot \sin \phi = r^2 \sin^2 \theta + r^2 \cos^2 \phi + r^2 \cos \phi \cdot \sin \theta$

Dividing through by $r^2$ and tidying up, we get $\cos 2\theta-\cos 2\phi=\sin(\theta+\phi)$

With a bit of manipulation, this becomes $\left(2\sin(\theta-\phi)+1 \right)\sin(\theta+\phi)=0$

Since $0<\theta+\phi<\pi$ , we can't have $\sin(\theta+\phi)=0$ ; therefore $2\sin(\theta-\phi)+1=0$

and $\phi=\theta+\frac{\pi}{6}$

The quantity we want to find is $\begin{aligned} x&=\frac{ab+cd}{ad+bc} \\ &=\frac{\sin 2\theta+\sin 2\phi}{2\sin(\theta+\phi)} \\ &=\frac{\sin 2\theta+\sin \left(2\theta+\frac{\pi}{3}\right)}{2\sin \left(2\theta+\frac{\pi}{6}\right)} \\ &=\frac{\sin 2\theta + \frac{\sqrt3}{2}\cos 2\theta + \frac12 \sin 2\theta}{\cos 2\theta + \sqrt3 \sin 2\theta} \\ &=\boxed{\frac{\sqrt3}{2}} \end{aligned}$