Sunday, May 26, 2019

Geometry Problem 1436: Right Triangle, Altitude, Excircles, Excenters, Geometric Mean, Art, Poster

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

Details: Click on the figure below.

Geometry Problem 1436: Right Triangle, Altitude, Excircles, Excenters, Geometric Mean, Art, Poster, iPad apps, Tutoring.
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2 comments:

  1. AnonymousMay 26, 2019 at 9:40 AM

    connect AD, EC,
    angle AHD =CHE =45 degree;
    angle DAH =45+ angle ACB/2 ;
    angle CEH = 180-45-angle ECH=135 - (45+angle BAC/2)= 90- angle BAC/2 =90 -(90-ACB)/2=45+angle ACB/2;
    triangle ADH similar to CEH,
    HD*HE=AH*HC=BH^2

    ReplyDelete
  2. Peter TranMay 26, 2019 at 4:06 PM

    https://photos.app.goo.gl/GRVbRNHhU7pWGEGf6

    Connect AD and CE
    Let a= ∠ (BAC)
    In right triangle ABC we have ∠ (HBC)=a
    Note that ∠ (EHC)= ∠ (AHD)=45
    And ∠ (DAH)= (180-a)/2= 90-a/2
    With angles manipulation we also get ∠ (HEC)= 90-a/2
    So triangle AHD similar to EHC ( case AA)
    And AH/EH= HD/HC => AH.HC= EH.HD
    In right triangle ABC with altitude BH we have the relation BH^2=HA x HC= EH x HD

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