Geometry Problem

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

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Complete Problem 444

Level: High School, SAT Prep, College geometry

## Friday, April 30, 2010

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

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

ReplyDeleteMethod 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'

extend AC to K, and AD to L ( K, L on circle O )

ReplyDeletejoin 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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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.

ReplyDeleteAP bisects both <CAD and <BAE, so AC and AD are isogonals.

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

DeleteYes, sorry abaut that mistake (:

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

regards.

A new solution with inversion!

ReplyDeleteLet 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.

Problem 444

ReplyDeleteI 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