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.