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?
 A: Noticing $\triangle{ADO}$ is an equilateral triangle where $O$ is the center of the circle should be a key.

Since the both sides of
$$\vert \sqrt{c+a}-\sqrt{c+b}\vert = \sqrt{c-a-b}$$
are non-negative, it is equivalent to
$$(\sqrt{c+a}-\sqrt{c+b})^2=c-a-b,$$
i.e.
$$c+2a+2b=2\sqrt{(c+a)(c+b)}$$
Since the both sides are positive, it is equivalent to
$$(c+2a+2b)^2=4(c+a)(c+b),$$
i.e.
$$\frac{3}{4}c^2=a^2+b^2+ab\tag1$$
So, all we need is to prove $(1)$.

Applying the law of cosines to $\triangle{ABC}$, we have
$$|\overline{AC}|^2=a^2+b^2-2ab\cos(120^\circ),$$
i.e.
$$|\overline{AC}|^2=a^2+b^2+ab\tag2$$
Applying the law of sines to $\triangle{ADC}$, we get
$$\frac{|\overline{CD}|}{\sin\angle{DAC}}=\frac{|\overline{AD}|}{\sin\angle{DCA}}$$
which implies
$$|\overline{AD}|=\frac c2$$
So, $\triangle{DAO}$ is an equilateral triangle where $O$ is the center of the circle. 
Applying the law of sines to $\triangle{ADC}$ gives
$$\frac{|\overline{AC}|}{\sin\angle{ADC}}=\frac{|\overline{AD}|}{\sin\angle{DCA}}$$ 
which implies
$$|\overline{AC}|^2=\frac 34c^2\tag3$$
Now, $(1)$ follows from $(2)(3)$.
A: You assigned the angle $\angle BCA = \theta$ and used trigonometry. 
You got the radius of the circle as $\frac{c}{2}$. 
Square 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. 


*

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

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


And reduced the equation to: $$1+2\sin\theta+2\sin(60^\circ - \theta)=2\sqrt{(1+\sin\theta)(1+\sin(60^\circ - \theta))}$$.
Reduce it using the addition-subtraction formulae of trigonometry.
$$1+2(\sin\theta+\sin(60-\theta) = 2\sqrt{(1+\sin\theta)(1+\sin(60-\theta)}$$
On reduction we get 
$$2m+1 =2\sqrt{m+m^2+\frac{1}{4}}$$
Where $m=\cos(30-\theta)$
On squaring and simplifying we get 
$$4m^2+4m+1=4m^2+4m+1$$ which is always true.
Hence the statement is proved. 
A: 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!
