Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
Use subscript 1,2,3 instead of a,b,c. Let BC=a, AC=b, AB=c. Let Δ be the area of ABC. r₁ = Δ/(s−a), 1/r₁ = (s−a)/Δ = (b+c−a)/(2Δ)r₂ = Δ/(s−b), 1/r₂ = (s−b)/Δ = (a+c−b)/(2Δ)r₃ = Δ/(s−c), 1/r₃ = (s−c)/Δ = (a+b−c)/(2Δ)ah₁ = 2Δ, 1/h₁ = a/(2Δ)bh₂ = 2Δ, 1/h₂ = b/(2Δ)ch₃ = 2Δ, 1/h₃ = c/(2Δ)The rest is obvious.
Let S= area of triangle ABC= Area(ACE)+Area(ABE)-Area(BCE)S= ½(.Ra).b+1/2(Ra).c-1/2(Ra).a= ½.(Ra).(b+c-a)So 1/(Ra)=1/2(b+c-a)/S ..... (1)But b/S=2/(Hb) and c/S=2/(Hc) , a/S=2/(Ha)Replace it in (1) we have 1/(Ra)=1/(Hb)+1/(Hc)-1/(Ha)