Problem on Law of Sines from 'Geometry Revisited' This problem is from 'Geometry Revisited', Exercise number 4 in section 1.1. The question and answer below are from the textbook.  Copy of book found here
Question:-

Let $p$ and $q$ be the radii of two circles through $A$, touching $BC$ at $B$ and $C$,
      respectively. Then $pq = R^2$.

Answer from the book :-

$$c = \frac{2p}{\sin B} = \frac{pb}{R},  
    b = \frac{2q}{\sin C} = \frac{qc}{R}$$ 

Multiply and simplify.

My understanding :-


*

*$R$ is the circumradius of triangle $ABC$. By Law Of Sines, $a/\sin A = b/\sin B = c/\sin C = 2R$.

*The side $AB$ is common to the 2 triangles having radii p and R, just
as $AC$ is common to other 2 triangles having radii q and R.

*By the Law Of Sines, $2p = c/\sin C'$ ,if we call C' as the side
which is the 3rd side of triangle whose radius is p and shares AB
with ABC.

*Similarly $2q = a/\sin A'$ if we call A' as the side which is not
shared by the triangle whose radius is q.


Have I understood correctly so far ? If then, how is $c=2p\sin B$?
 A: Let $AB'$ a diameter of the circle whose radius is $p$ and $AC'$ a diameter of the circle whose radius is $q$.
See the following figure:

From $\triangle ABB'$ we get:
$$\sin B'= \frac{c}{2p}, \quad (1)$$
and from $\triangle ACC'$:
$$\sin C'= \frac{b}{2q}. \quad (2)$$
But $B'=B$ and $C'=C$ (Why?).
Recall that:
$$\sin B= \frac{b}{2R}, \quad (3)$$
and
$$\sin C= \frac{c}{2R}. \quad (4)$$
From $(1)$ and $(3)$ we get:
$$c=\frac{pb}{R}, \quad(5)$$
and 
and from $(2)$ and $(4)$ we get:
$$b=\frac{qc}{R}. \quad(6)$$
Finally from $(5)$ and $(6)$ we get:
$$pq=R^2$$
A: I have a slightly different solution that doesn't require the inscribed angle theorem.
First, I actually think this question is a bit ambiguous and confusing. For more on that see:
Problem 1.1.4 from "Geometry Revisited"
In short, the statement is true if "$BC$" is interpreted as "line $BC$" and false if interpreted as "line segment $BC$".
Let $P$ be the midpoint of the circle with radius $p$. Since this circle touches the line $BC$ in exactly one place, which is $B$, the angle $CBP$ is a right angle. Thus the angle $ABP$ measures $\pi/2 - B$ radians. Since the triangle $ABP$ is isosceles, the rest of the angles can be computed. Applying law of sines to the triangle and using $\sin(2B) = 2 \sin(B)\cos(B)$ gives the desired result $c = 2p \sin(B)$.
Edit: The above argument holds only if angle $ABC$ is acute. If it is obtuse, then a similar argument shows that the angle $ABP$ measures $B-\pi/2$ and the rest of the argument is similar. If angle $ABC$ is a right angle, then $c=2p \sin(B)$ follows directly.
