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. 

$$\angle AOD=2\angle ACD$$
$$\angle AOC=2\angle ADC$$
$$\angle BOD=2\angle BAD$$
$$\angle BOC=2\angle BAC$$
$$\angle ACD+\angle BAC=\angle BAD+\angle ADC=90^{\circ}$$
$$\therefore \quad \angle ACD-\angle ADC=\angle BAD-\angle BAC$$
$$\therefore \quad \angle AOD-\angle AOC=\angle BOD-\angle BOC$$
$$\therefore \quad \operatorname{arc} AD-\operatorname{arc} AC
= \operatorname{arc} BD-\operatorname{arc} BC$$

