Let ABC be a triangle. Let [math]p[/math] and [math]q[/math] be the radii of two circles through A, touching BC at B and C, respectively. Prove that [math]pq=r^2[/math], where [math]r[/math] is the radius of the circumcircle of ABC. (Source: Geometry Revisited by Coxeter & Greitzer, 1967, p3)
1. Visualise the problem by moving F & G so that both FB and GC are perpendicular to BC. 2. Observe the numerical results. 3. Repeat 1 & 2 for various triangles, ABC. 4. What construction(s) might be helpful to prove the desired result? 5. Hint: Use the extended Law of Sines, [math]a[/math]/sin A = [math]b[/math]/sin B = [math]c[/math]/sin C = [math]2r[/math].