Finding the common face of clinging soap bubbles using trigonometric functions of angles I am trying to help my daughter with a problem from Stewart's Precalculus book.This problem comes right after law of sines.
When two bubbles cling together in midair, their common surface is part of a sphere whose center D lies on the line passing through the centers of the bubbles (please refer to the figure below) also angles ACB and ACD each have measure 60 degrees


*

*Show that the radius r of the common surface is given by r = ab / (b - a)

*Find the radius of the common face if the radii of the bubbles are 3cm and 4cm


I could do the second one but after using law of cosines to find length of the segment AB in triangle CBA. That came out as 
Then I used law of sines in triangle ABC to find angle CAB = 73.897 degrees
Angle CAD = 180 - angle CAB = 106.1 degrees
angle CDA = 180 - 106.1 - 60 = 13.897 degrees
Then I used law of sines in triangle CAD to find the value of r
But I couldn't make any headway for the first one. Also it seems to me that I don't need law of cosines to solve this problem.
Any help will be appreciated.
Thanks

 A: By the law of cosines,
$$BD=\sqrt{a^2+r^2-2ar\cos(120)}$$
$$BA=\sqrt{a^2+b^2-2ab\cos(60)}$$
$$AD=\sqrt{b^2+r^2-2br\cos(60)}$$
Since $BD=BA+AD$ we now have
$$\sqrt{a^2+r^2-2ar\cos(120)}=\sqrt{a^2+b^2-2ab\cos(60)}+\sqrt{b^2+r^2-2br\cos(60)}$$
Note that $\cos(60)=1/2$ and $\cos(120)=-1/2$. Hence we obtain
$$\sqrt{a^2+r^2+ar}=\sqrt{a^2+b^2-ab}+\sqrt{b^2+r^2-br}$$
WolframAlpha now gives the solution $r=ab/(a-b)$, although you can prove it by hand if necessary by squaring both sides, isolating the remaining root and then squaring both sides again.
For the second one, just plug in $a=4$ and $b=3$ to obtain $r=12$.
A: Hint.
The line $CA$ is angle $\angle{DCB}$ bisector so
$$
\frac{BA}{AD} = \frac{a}{r}
$$
A: As this is a homework problem, I will just give an outline of the answer. Denote the length of $\overline{AD}$ by $\ell$ and the length of $\overline{BA}$ by $s$.
Step 1: Find nice formulae for $\ell^2, s^2$ in (and at worst quadratic in) $a,b,r$ by using the law of cosines.
Step 2: Use the law of sines, and that $\sin( \angle BAC) = \sin(\angle DAC)$, to show $a/s = r/\ell$.
Step 3: The equation $r^2 = a^2 \ell^2/s^2$ is quadratic in $r$. Show that it has as solutions $r = a, ab/(a-b)$.
Step 4: Figure out why the solution $r=a$ is unnecessary.
