In given figure, prove In the given figure, two chords $AB$ and $CD$ intersect at right angles at $X$. Prove that $\text{arc } AD - \text{arc } CA =\text{arc }BD - \text{arc } BC$

My Attempt:
Join $AC$ and $AD$
$1. \quad \angle CAB+\angle ACD=\angle AXD$
$2. \quad \angle DAB +\angle ADC=\angle AXC$.
$3. \quad \angle AXC=90^{\circ}$
What should I do next?  Please complete the proof. 

Using this standard result:
$$\angle AXD = \frac{\operatorname{arc}AD+\operatorname{arc}BC}{2}$$
we also get
$$\angle AXC = \frac{\operatorname{arc}AC+\operatorname{arc}BD}{2}$$
but $\angle AXD=\angle AXC=90$, so
$$\frac{\operatorname{arc}AD+\operatorname{arc}BC}{2}=\frac{\operatorname{arc}AC+\operatorname{arc}BD}{2}$$
$$\operatorname{arc}AD+\operatorname{arc}BC=\operatorname{arc}AC+\operatorname{arc}BD \Rightarrow \operatorname{arc}AD-\operatorname{arc}AC=\operatorname{arc}BD-\operatorname{arc}BC$$
