Fallacy in showing that $\vert \sqrt{c+a}-\sqrt{c+b}\vert = \sqrt{c-a-b}$ Given a cyclic quadrilateral $ABCD$, $AB=a, BC=b, CD=c, \angle ABC=120^\circ, \angle ABD=30^\circ$, then, show that $$\vert \sqrt{c+a}-\sqrt{c+b}\vert = \sqrt{c-a-b}$$
I tried to do it using some trig bashing. 
What I did was basically assigned the angle $\angle BCA = \theta$ and used trigonometry. 
It's not hard to see that the radius of the circle is $\frac{c}{2}$. 
I've squared the equation on both sides and obtained: $$c+2a+2b=2\sqrt{(c+a)(c+b)}$$. 
And using $\text {Sine Rule}$, the equation can be reduced further. 
I used 


*

*$\frac{a}{\sin\theta}=c$

*$\frac{b}{\sin(60^\circ - \theta)}=c$


And reduced the equation to: $$1+2\sin\theta+2\sin(60^\circ - \theta)=2\sqrt{(1+\sin\theta)(1+\sin(60^\circ - \theta))}$$.
Then I continued to reduce it using the addition-subtraction formulae of trigonometry.
And at the end of the day what I get is $$\boxed{\cos(30^\circ-\theta)=\frac{1}{2}}$$ after all those addition-subtraction of trigonometric equations.
And that's not true I believe.
May I get rectified?

By law of cosines $$AC=\sqrt{a^2+b^2-2ab\cos120^{\circ}}=\sqrt{a^2+b^2+ab}.$$
In another hand, since $\measuredangle CBD=90^{\circ},$ we see that $CD$ is a diameter of the circle.
Thus, $$\cos30^{\circ}=\frac{\sqrt{a^2+b^2+ab}}{c}$$ or
$$3c^2=4(a^2+b^2+ab),$$ which is $\vert \sqrt{c+a}-\sqrt{c+b}\vert = \sqrt{c-a-b}$ exactly!
