Thread: Another proof with Law of Sines

1. 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.

2. Re: Another proof with Law of Sines

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.

3. Re: Another proof with Law of Sines

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

4. Re: Another proof with Law of Sines

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.