Sunday, July 15, 2012

Problem 782: Triangle, Orthocenter, Circumcenter, Circumcircle, Diameter, Altitude, Parallelogram

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

Click the figure below to see the complete problem 782.

Online Geometry Problem 782: Triangle, Orthocenter, Circumcenter, Circumcircle, Diameter, Altitude, Parallelogram.

Posted by Antonio Gutierrez at 6:32 PM
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4 comments:

  1. Jacob HaJuly 15, 2012 at 7:01 PM

    AD diameter => ABD=90 degrees
    => BD//HC

    AD diameter => ACD=90 degrees
    => DC//BH

    Hence, BDCH is parallelogram.

    ReplyDelete
  2. W FungJuly 15, 2012 at 7:14 PM

    let the altitudes from A,B,C touches the opposite sides at P,Q,R.
    Angle PHC = Angle ABC = Angle ADC
    Angle BHP = Angle ACB = Angle BDA
    So Angle BHC = Angle BHP + Angle BHP = Angle ADC + Angle BDA = Angle BDC

    Angle ABQ = Angle ACR
    Hence, Angle HBD = 90 - Angle ABQ = 90 - Angle ACR = Angle HCD

    So we have proved two same pairs of opposite angles.
    Q.E.D.

    ReplyDelete
  3. W FungJuly 15, 2012 at 10:19 PM

    (Alternative way by Vector)
    let vector OA,OB,OC be a,b,c respectively.
    Then vector OH = a+b+c, OD = -a
    vector BD = OD - OB = -a-b
    vector HC = OC - OH = c - (a+b+c) = -a-b
    Hence vector BD = vector HC
    A parallelogram is formed
    q.e.d.

    ReplyDelete
  4. PravinJuly 15, 2012 at 11:21 PM

    Let M be the midpoint of AB
    CH // OM // BD and
    CH = 2 OM = BD
    Follows BDCH is a parallelogram

    ReplyDelete

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