Wednesday, October 7, 2015

Geometry Problem 1150: Circle, Perpendicular Secants, 90 Degrees, Congruence

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

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Online Math: Geometry Problem 1150: Circle, Perpendicular Secants, 90 Degrees, Congruence.

Posted by Antonio Gutierrez at 3:48 PM
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7 comments:

  1. Sumith PeirisOctober 8, 2015 at 3:11 AM

    Extend AH to M which is on BE. < MEA = < ACD = < EAM = < GCA = < DCM since Tr. ACD & MCD are therefore congruent. So 2AH = AM = ME = BM = AG

    Now I need to show that BM = GF or that GB = FM which step eludes me still. Dear Antonio could u help?

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
    Replies
    1. Sumith PeirisOctober 8, 2015 at 7:51 AM

      Am I on the right track Antonio?

      Delete
    2. Antonio GutierrezOctober 8, 2015 at 9:37 AM

      Dear Sumith,
      In your first statement < MEA = < ACD = < EAM = < GCA = < DCM
      why <DCM = < GCA = ....

      Thanks
      Antonio

      Delete
  2. c.t.e.o.October 8, 2015 at 10:40 AM

    Extend CG and EA let be P point where them meet each other. Draw MN //GF9Through A). Triangles PMA kongruent to AHD(PCD isoceles) PMA kongruent to ADN => MA=AN=AH

    ReplyDelete
  3. Stan FulgerOctober 19, 2015 at 1:04 PM

    CBDE is cyclic, so <BEA=<ACD (1), CGAE is cyclic, so <BEA=<BCG (2), hence BC is bisector of <GCD (3). Let M be the intersection of AD and CG, N the intersection of BC and DF; from (3) and AD_|_BC, CMND is a rhombus. GF is one of its altitudes, and AH is half of another altitude, done.

    ReplyDelete
  4. Ivan BazarovNovember 24, 2015 at 5:31 PM

    Because CBDE is cyclic, a=<BED=<BCD
    Because CGAE is cyclic, <BED=<GCB=a
    Therefore, <GCD=2a.
    GF=CD*sin(2a)=2CDcos(a)sin(a)=2cos(a)(CDsin(a))=2cos(a)AD.
    Note that from similar triangles, <HAD=<ACD=a, so ADcos(a)=AH.
    Plugging this into previous equation gives GF=2cos(a)AD=2AH.

    ReplyDelete

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