Right Triangle ProofDate: 11/19/2004 at 10:26:03 From: Kaustav Subject: Inscribed Circle In right triangle ABC, let CD be the altitude to the hypotenuse. If r1,r2,r3 are radii of the incircles of triangles ABC, ADC, and BDC, respectively, prove CD = r1 + r2 + r3. I'm not sure how to approach it. I have considered using areas. If I can prove the ratio of radii to the sides, I think it can be solved. Any hints? Date: 11/19/2004 at 12:36:46 From: Doctor Schwa Subject: Re: Inscribed Circle Hi Kaustav, By cutting each triangle into three, you can prove that the area of each triangle = 1/2 * radius of inscribed circle * perimeter of triangle. So: Area ABC = 1/2 * r1 * (AB + BC + AC) Area ADC = 1/2 * r2 * (AD + DC + AC) Area BDC = 1/2 * r3 * (BD + DC + BC) and since area ABC = area ADC + area BDC, r1*AB + r1*BC + r1*AC = r2*AD + r2*DC + r2*AC + r3*BD + r3*DC + r3*BC which is some progress, since now we know DC = (r1*AB + r1*BC + r1*AC - r2*AD - r2*AC - r3*BD - r3*BC)/(r2 + r3) but this still looks pretty ugly. In my argument, though, I never used the fact that the triangle was right! So, what else do we get? Area ABC = 1/2 * AC * BC and also some nice proportions like AD/DC = AC/CB = DC/DB. There's some other nice facts, too, like because tangent segments to a circle must be equal, r1 = (AC + BC - AB)/2 (do you see how to prove this fact?) and analogous relationships in the smaller triangles as well. I think this last idea is VERY promising ... try writing down the expressions for r2 and r3 that you get by this same method and see if you can find that r1 + r2 + r3 = CD! - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
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