## Monday, May 19, 2008

### Elearn Geometry Problem 36

See complete Problem 36
Right triangle, altitude, incircles. Level: High School, SAT Prep, College geometry

1. We've BD = ca/b, AD = c^2/b and therefore,
r1=(ac/b+c^2/b)/(c+ca/b+c^2/b)= ac^2/(b(a+b+c))
Likewise, r2 = ca^2/(b(a+b+c)) and r=ac/(a+b+c)
Hence, LHS = ac^3/(b(a+b+c))+ca^3/(b(a+b+c))
= ac(c^2+a^2)/(b(a+b+c))
= acb/(a+b+c) = r*b since r=ac/(a+b+c)
Thus, r1*c + r2*a = r*b or r1*AB+r2*BC=r*AC
QED
Ajit: ajitathle@gmail.com

2. As ADB, BDC, and ABC are similar, AB/r1=BC/r2=AC/r=k, so AB=kr1, BC=kr2, AC=kr.
From AB^2+BC^2=AC^2 we get AB.kr1+BC.kr2=AC.kr and finaly r1.AB+r2.BC=r.AC.

3. Our proof is based on the simple fact that the in-radius of a triangle XYZ right angled at Z is given by (XZ+YZ-XY)/2.
Accordingly,
r1 = (AD + DB - AB)/2 and r = (AB + BC - AC)/2
So r1/r = (AD + DB - AB)/(AB + BC - AC)
= (AB/AC).[AD/AB + DB/AB - 1]/[AB/AC + BC/AC - 1]
= cos A.(cos A + sin A - 1}/(cos A + sin A - 1}
= cos A
Similarly r2/r = cos C
Hence
r1.AB + r2.BC
= r1.c + r2.a
=(r cos A)c + (r cos C)a
= r(a cos C + c cos A)= r.b
= r.AC

4. By similar triangles ADB and ABC we have