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 670.
http://img835.imageshack.us/img835/1895/problem670.pngPer the result of problem 662We have ( AHC)=(AIC)=(AOC)=120And 5 points A, H, I, O and C are co cyclic ( see picture)Since (B)=60 >> (BAH)=30Since (AOC)=120 >> (OAC)=30AI is and angle bisector of (BAC) >> AI is also an angle bisector of (HAO)Arc HI=arc IO and IH=IOPeter Tran
B = 60°, A + C = 120°. Let D be the midpoint of BCBH = 2 OD = 2R cos B = R = BO∠HBA = 90° - A ∠OBC = ∠OCB = 90° - ∠BOC/2 = 90° - A = ∠HBAAlready BI bisects ∠ABCFollows BI bisects ∠HBO and ∆BIH ≡ ∆BIO∴HI = IO
Let D be midpoint of BC and let CH extended meet AB at E.BE = a/2 = BD and < ABH = < OBD = 90-AHence tr.s BHE and BDO are congruent ASA and so BH = BO.Now < HBI = < OBI = B/2 - (90-A)Hence tr.s HBI and OBI are congruent SASTherefore HI = OISumith PeirisMoratuwaSri Lanka