Online Geometry theorems, problems, solutions, and related topics.
See complete Problem 36Right triangle, altitude, incircles. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
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*ACQEDAjit: firstname.lastname@example.org
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.
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)/2So 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 ASimilarly r2/r = cos CHence 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