Wednesday, January 4, 2012

Problem 713: Tangent Circles, Diameter, Chord, Tangent, Incenter, Incircle, Triangle

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 713 details.

Online Geometry Problem 713: Tangent Circles, Diameter, Chord, Tangent, Incenter, Incircle, Triangle.

2 comments:

  1. http://img12.imageshack.us/img12/9500/problem713.png
    Connect 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

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  2. Referring to Problem 712, recall that
    x = (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.

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