Online Geometry theorems, problems, solutions, and related topics.
Proposed ProblemSee complete Problem 23 at:gogeometry.com/problem/p023_right_triangle_incircles.htmLevel: High School, SAT Prep, College geometry
Under the standard nomenclature, AD = c*sin(C)= c*c/b=c^2/b while BD=c*cos(C)= ac/b ----(1)So r1 = AD*BD/(AD+BD+AB) = (c^2/b*ac/b)/(c^2/b + ac/b + c) = ac^2/b(a+b+c)Similarly, r2 = ca^2/b(a+b+c)and r = ac/(a+b+c)So,r1+r2+r=ac^2/b(a+b+c)+ca^2/b(a+b+c)+ac/(a+b+c) = ac/b = BD by equation (1)Ajit: firstname.lastname@example.org
The statement that r=ac/(a+b+c) is incorrect; the numerator should be 2ac.However, it is easier to show that r=a+b-c and use that to obtain the desired result.
Prof Elie Achkar (Lebanon)BD + DC = BC + 2r1 in the triangle BDCAD + BD = AB + 2r2 in the triangle BDAAB + BC = AC + 2r in the triangle ABCthen :BD+DC+AD+BD+AB+BC = 2r1 + 2r2 + 2rTHEN :2BD = 2(r1 + r2 +r )follow : BD = r1 + r2 + r
Inradius = Tr. Area/Semisum = (ac/2)/[(a+b+c)/2]Hence r = ac/(a+b+c) So then why is the statement incorrect?
Dear Joe = Ajit:I agree, your statement r = ac/(a+b+c) is correct.
How is it possible thatBD + DC =BC + 2r1..................................Please tell me the condition no
In a right triangle ABC (right angled at B) r=s–b,is well-known.Easily remembered as : in-radius = semi perimeter–hypotenuse (or)in-diameter=perimeter–twice the hypotenuseSo 2(r + r1 + r₂) =(AB+BC–AC)+(BD+AD–AB)+(BD+DC–BC)=2BD+(AD+DC)–AC=2BD+AC–AC =2BDHence r+r1+r₂=BD
If O1, O2 and O are the centers of circles with radius r1,r2 and r then r1 = rc/b and r2 = ra/b since Tr.s AO1C, CO2B and AOC are similar So r+r1+r2 = r (a+b+c)/b ....(1)By writing the area of the right Tr. in 2 waysBD. b = r(a+b+c)So from (1) r+r1+r2 = BDSumith PeirisMoratuwaSri Lanka