Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the problem 713 details.
Wednesday, January 4, 2012
Problem 713: Tangent Circles, Diameter, Chord, Tangent, Incenter, Incircle, Triangle
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http://img12.imageshack.us/img12/9500/problem713.png
ReplyDeleteConnect O’D, BC and IC
We have O’D perpen. AC and BC perpen. To AC
Let 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 isosceles
Angle (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 that
ReplyDeletex = (90° + 25°)/2,
more generally, if we denote ∠CAB by θ we can
prove similarly that x = (90° + θ)/2
Applying it to the the present Problem 713,
∠CIB = (90° + θ)/2
Follows that ∠ICE = (90° - θ)/2 = ∠ACE /2
So CI bisects ∠ACE.
Already AB bisects ∠CAE
Hence I is the incentre of ∆ACE.