Monday, July 8, 2013

Problem 898: Intersecting Circles, Common External Tangent, Secant, Circumcenter, Collinear Points

Geometry Problem. GeoGebra, HTML5 Animation for iPad and Nexus.
Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the dynamic geometry demonstration of problem 898.

Online Geometry Problem 898: Intersecting Circles, Common External Tangent, Secant, Circumcenter, Collinear Points. GeoGebra, HTML5 Animation for iPad and more tablets

Posted by Antonio Gutierrez at 4:18 PM
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1 comment:

  1. Peter TranJuly 8, 2013 at 5:10 PM

    Power of H to circle O= HC^2=HA.HB
    Power of H to circle O1=HD^2=HA.HB
    So HC=HD => H is the midpoint of CD
    Per result of problem 897, CGD is an isosceles triangle
    So HG is the perpen. bisector of CD and H, O2 and G are collinear

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