Right Triangle Proof

Date: 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/