Thread: Another proof with Law of Sines

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.

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 (the angle at A), (the angle at B), and (the angle at C). Show that the angle BOC is the angle BPA is and the angle AQC is Deduce that and Then use the sine rule to get the result.

How do I find that 2A = BOC, and so on

Originally Posted by scruz10

How do I find that 2A = BOC, and so on

Circle theorems: (1) angles at centre and circumference; (2) alternate segment theorem.