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