Friday, July 20, 2012

Problem 788: Intersecting Circles, Secant Line, Midpoint, Cyclic Quadrilateral

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 788.

Online Geometry Problem 788: Intersecting Circles, Secant Line, Midpoint, Cyclic Quadrilateral.

Posted by Antonio Gutierrez at 2:21 PM
Labels: , ,

6 comments:

  1. Peter TranJuly 20, 2012 at 3:55 PM

    http://img687.imageshack.us/img687/6215/problem788.png
    Draw lines per attached sketch
    We have <BDA=<BFA=.5*<BQA=<BQO
    And <BEA=<BCA=.5*<BOA=<BOQ
    So all 3 triangles OBQ, EBF and CBD are similar per case AA
    Tri. CBD is the image of Tri. EBF in the spiral similarity transformation centered B , angle of rotation = theta=<EBC=<FBD=<FAD
    Since G is the image of H in this transformation so <HBG=<FAD= theta
    So quadrilateral AHBG is cyclic

    ReplyDelete
  2. Jacob HaJuly 20, 2012 at 5:01 PM

    Join BC, BD, BE and BF.

    Angle AFB = angle ADB
    Angle AEB = angle ACB

    Thus, triangle BCD ~ triangle BEF.

    Now rotate triangle BCD to triangle BEF,
    with center B, such that
    BC and BE coinside, BD and BF coinside,
    with different lengths.

    Then
    CD becomes EF (with different lengths),
    BG becomes BH (with different lengths).

    Therefore, the rotation angle is
    angle GBH, and also same as angle CAE.

    Hence, AHBG is concyclic.

    ReplyDelete
  3. PravinJuly 21, 2012 at 8:58 AM

    Let M be the midpoint of OQ. Join OA, QA, MB
    Power of H w.r.t. circle (O) = HA.HE = OA² - OH²
    Power of H w.r.t. circle (Q)= HA.HF = HQ² - QA²
    So OA² - OH² = HQ² - QA² (since HE = HF),
    OA² + QA² = OH² + HQ² which is same as
    2AM² + MQ²/2 = 2HM² + MQ²/2
    Follows MA = MH
    Similarly equating powers of G w.r.t. circle (Q),it can be shown that MA = MG
    OQ being the perpendicular bisector of AB,we have MA = MB
    Hence MA = MH = MG = MB and A, H, B, G are concyclic.

    ReplyDelete
  4. Sumith PeirisApril 25, 2016 at 4:43 AM

    Let CG = GD = p and let EH = HF = q

    Tr.s BCD and BFE are similar so,

    2p/2q = s/t where BC = s and BE = t

    Hence s/p = t/q

    So in Tr. s BEH and BCG

    < E = < C and the sides which hold these angles are proportional

    Hence

    ReplyDelete
  5. Sumith PeirisApril 25, 2016 at 4:43 AM

    ...the triangles are similar and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  6. APOSTOLIS MANOLOUDISAugust 5, 2016 at 9:52 AM

    Problem 788
    Is <ACB=<AEB,<ADB=<AFB then triangleCBD similar triangleEBF.So BD/BF=CD/EF=2GD/2HF=GD/HF.Then triangleBDG similar triangleBFH .
    So <BGD=<BHF=<BHA.Therefore AHBG is cyclic.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete

Subscribe to: Post Comments (Atom)