## Friday, April 30, 2010

### Problem 444: Tangent circles, Secant line, Chords, Angles, Congruence

Geometry Problem
Click the figure below to see the complete problem 444 about Internally Tangent circles, Secant line, Chords, Angles, Congruence.

Complete Problem 444

Level: High School, SAT Prep, College geometry

1. Let AB,AE intercept circle O' at B',E'

Method 1:
mAEB=mAAB=mACB' (can also be explained by homothety)
mEDA=mCB'A
thus α=α'

Method 2:
consider homothety, B'E'//BE => B'C=E'D => α=α'

2. extend AC to K, and AD to L ( K, L on circle O )
join K to L
join A and E to O', O
join C to P ( P point AE meet circle O' )
=>
AO'P = AOE (1) (AO'P and AOE isoc and OAE common)
ACP = 1/2 AO'P
AKE = 1/2 AOE
from (1)
ACP = AKE (2)

PCD = DAP = EKL = α' (3) ( have same arc EL, PD)
from (2) and (3)
ACE = AKL
=>
BE // KL
=>
BEK = α = EKL = α' as alternate angles
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3. A lot simplier: Let T be a tangent line to smaller circle wich is parallel to AB and let P be his touchng point with this circle.
AP bisects both <CAD and <BAE, so AC and AD are isogonals.

1. To Editor (problem 444), in your solution T should be parallel to BE

2. Yes, sorry abaut that mistake (:

PD: I wonder why Blogger changed mi name randomly to "Editor", so far was "mathreyes"

regards.

4. A new solution with inversion!

Let us take point A as pole and with any radius invert de figure.
You will see without problems that the inverted problem is "Let us consider PQR a triangle and K its circumcircle, take a point U in arc QR not containing A such that the tangent line to K through U is paralel to QR. Prove that AU is bisector of <A" witch si obviously true.

5. Problem 444
I design the common external tangent at point A xAx’.Then <xAB=<AEB, <xAC=<ADC=<AED+
<DAE so <BAC=<DAE.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE