Another proof with Law of Sines

**scruz10** · Aug 29th 2011, 02:58 PM

Let p and q be the radii of two circles through A (with A being a point on Triangle ABC), touching BC at B and C, respectively. Prove the pq = R squared. (R is the radius of the circle circumscribed around ABC).

I have been trying over and over to do this problem, and I just can't get it.

**Opalg** · Aug 30th 2011, 05:39 AM

Originally Posted by scruz10

Let p and q be the radii of two circles through A (with A being a point on Triangle ABC), touching BC at B and C, respectively. Prove the pq = R squared. (R is the radius of the circle circumscribed around ABC).

I have been trying over and over to do this problem, and I just can't get it.

Let O be the centre of the circumcircle, and P, Q the centres of the other two circles, as in the picture. Let the angles of the triangle be $\alpha$ (the angle at A), $\beta$ (the angle at B), and $\gamma$ (the angle at C). Show that the angle BOC is $2\alpha,$ the angle BPA is $2\beta,$ and the angle AQC is $2\gamma.$ Deduce that $BC = 2R\sin\alpha,$ $AB = 2p\sin\beta,$ and $AC = 2q\sin\gamma.$ Then use the sine rule to get the result.