Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 668.
http://img819.imageshack.us/img819/5083/problem668.pngConnect CO and AONote that CO and CO2 are angle bisector of angle (BCA) soC, O, O2 are collinearTriangle OEC congruence with triangle OHC ( right triangles with common hypotenuse and CO is angle bisector)So OC is the angle bisector of angle (HOE) >>Incircle O2 tangent to OE also tangent to OH.Similarly AO is angle bisector of angle HOF) and incircle O1 tangent to both OF and OHPeter Tran
Let circle O₁ touch AB at J, AC at K & FG at L.Complete the square O₁JHMO₁M = KH = AH - AK = AF - AJ = KF = radius of O₁ (note FJ O₁K is a square)follows OH touches circle O₁.Similarly we can show OH touches circle O₂
Refer my solution to Problem 668:"FJO₁L is a square" (not FJO₁K as previously typed)
another point of vuein ABC r=s-btriangles AFG and ABC are homothetic,center A ratio (s-a)/cthe incircle of AFG is the image of the incircle of ABC;r1=(s-a)r/c=(s-b)(s-a)/cline OH becomes the parallel line O1H1distance H1H=(s-a)-(s-a)²/c=(s-a)(s-b)c=r1incircle O1 tangent to OHThe same way CDE and ABC are homothetic with center C and ratio (s-c)/a