Prove this geometrical statement Let ${p}$ and ${q}$ be the radii of two circles passing through ${A}$,touching ${BC}$ at ${B}$ and ${C}$, respectively. Then prove that ${p.q=R^2}$ where ${R}$ is the circumradius of ${ABC}$.
Thank you! :)
Edited: @newbie : This is the construction according to your details! 
                  Its not coming accurate!

The question represents the following figure.

One has to prove that ${\LARGE{p.q=R^2}}$.
My Thoughts: I am thinking of the following logic(2nd diagram):
${{\LARGE{cos}}(\angle{ABE})=\frac{\frac{c}{2}}{p}}$     ...(1)
${{\LARGE{cos}}(\angle{ACG})=\frac{\frac{b}{2}}{q}}$     ...(2)
And from Extended Law of Sines we get,
${{\LARGE{sin}}(\angle{ACB})=\frac{c}{2R}}$       ...(3)
${{\LARGE{sin}}(\angle{ABC})=\frac{b}{2R}}$       ...(4)
Now using these four equations, I have to prove the result! This is where I'm stuck and need help.
 A: I assume here that $\overleftrightarrow{BC}$ should be tangent to the circles, whose centers I denote $P$ and $Q$. Let $R$ be the center of the circumcircle of $\triangle ABC$, and let the circumradius be $r$.

Write $\beta$ and $\gamma$ for $\angle ABC$ and $\angle ACB$, respectively. Noting that $\angle PBC$ and $\angle QCB$ are right angles (by the tangency of $\overleftrightarrow{BC}$), we can deduce that $\angle APB = 2\beta$ and $\angle AQC = 2\gamma$. (This is actually an aspect of the Inscribed Angle Theorem.) Consequently, 
$$|\overline{AB}| = 2p\;\sin\beta \qquad \text{and} \qquad |\overline{AC}| = 2q\;\sin\gamma$$
By the Law of Sines, we know that 
$$2 r = \frac{|\overline{AC}|}{\sin\beta} = \frac{|\overline{AB}|}{\sin\gamma}$$
whence
$$(2 r)^2 = \frac{|\overline{AC}|}{\sin\beta} \cdot \frac{|\overline{AB}|}{\sin\gamma} = \frac{2q\;\sin\gamma}{\sin\beta} \cdot \frac{2p\;\sin\beta}{\sin\gamma} = 4 p q \qquad \to \qquad r^2 = p q$$
Note that the argument holds regardless of which point of intersection of $\bigcirc P$ and $\bigcirc Q$ is taken to be $A$.
A: Draw lines $BE$, $CF$, $AB$, $AC$ and $EF$ ($E$ and $F$ are the centers of two circles respectively).
Furthermore, draw a line orthogonal to $AB$ intersecting $BC$ at $G$ and link $AG$. Because $\triangle BEG\cong\triangle AEG$ we know $AG\perp EF$ and $AG=BG$. Similarly we have $AG = CG$.
On the other hand, since $\angle BEA+ \angle AFC = \pi\implies \angle EAB + \angle AFC = \pi/2\implies\angle BAC = \pi/2$.
Once again we have $AG = BG = CG$, given $BG = CG$. Furthermore they are the circumradius of $ABC$.
The equality $pq = R^2$ can be obtained from the similarity of e.g. $\triangle EGA$ and $\triangle EGF$.
