Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 713 details.
http://img12.imageshack.us/img12/9500/problem713.pngConnect O’D, BC and ICWe have O’D perpen. AC and BC perpen. To ACLet H is the projection of O’ over BC (see picture)DCHO’ is a rectangle and tri. IDO’ congruence to tri. HO’B ----(case ASA)So O’H=DC=DI >>> Tri. IDC is isoscelesAngle (DIC)=(ICE)=(DCI)----- (alternate angle)So CI is an angle bisector of tri. ACE and I will be incenter of tri. ACE
Referring to Problem 712, recall thatx = (90° + 25°)/2,more generally, if we denote ∠CAB by θ we can prove similarly that x = (90° + θ)/2Applying it to the the present Problem 713,∠CIB = (90° + θ)/2 Follows that ∠ICE = (90° - θ)/2 = ∠ACE /2So CI bisects ∠ACE.Already AB bisects ∠CAEHence I is the incentre of ∆ACE.