Advanced Geometric Problem & Solution (Twice a week update) Now 4565 problems posted.
<Tri-2024-2129>, hint
Triangles ABC and AED are similar and BC/ED=r/(r(a)
2[ABC]=r(a+b+c)=a h(a)
r(a)/r=(h(a)-2r)/h(a).=1-2r/h(a)=1-2 a/(a+b+c)
(r(a)+r(b)+r(c))/r=3-2(a+b+c)/(a+b+c)=1.
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