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Thread: Another proof with Law of Sines

  1. #1
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    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.
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  2. #2
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    Re: Another proof with Law of Sines

    Quote Originally Posted by scruz10 View Post
    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 $\displaystyle \alpha$ (the angle at A), $\displaystyle \beta$ (the angle at B), and $\displaystyle \gamma$ (the angle at C). Show that the angle BOC is $\displaystyle 2\alpha,$ the angle BPA is $\displaystyle 2\beta,$ and the angle AQC is $\displaystyle 2\gamma.$ Deduce that $\displaystyle BC = 2R\sin\alpha,$ $\displaystyle AB = 2p\sin\beta,$ and $\displaystyle AC = 2q\sin\gamma.$ Then use the sine rule to get the result.
    Attached Thumbnails Attached Thumbnails Another proof with Law of Sines-circles.png  
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    Re: Another proof with Law of Sines

    How do I find that 2A = BOC, and so on
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    Re: Another proof with Law of Sines

    Quote Originally Posted by scruz10 View Post
    How do I find that 2A = BOC, and so on
    Circle theorems: (1) angles at centre and circumference; (2) alternate segment theorem.
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