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.

Problem 444: Internally Tangent circles, Secant line, Chords, Angles, Congruence
See also:
Complete Problem 444

Level: High School, SAT Prep, College geometry

7 comments:

  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'

    ReplyDelete
  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
    -----------------------------------------------

    ReplyDelete
  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.

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

      Delete
    2. Yes, sorry abaut that mistake (:

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

      regards.

      Delete
  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.

    ReplyDelete
  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

    ReplyDelete