Interactive step-by-step animation using GeoGebra. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Details: Click on the figure below.
Monday, April 27, 2020
Dynamic Geometry 1475: Clifford Intersecting Circles Theorem, Step-by-step Illustration
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https://photos.app.goo.gl/oeaR3NCro86whAueA
ReplyDeleteDefine points A,B,C,D,E,F as shown on the sketch
Perform geometric inversion with center at P and circle power k
= PP34. PA=PP24. PD=PP14. PE=PP12. PC= PP23. PF=PP13. PB
In this inversion ,Green circles C1 , C2, C3 and C4 will become
Lines BCE, DCF, ABF, and ADE.
And red circles C123, C124, C234 and C134 will become
Circumcircles of triangles FBC, CDE, ADF and ABE.
Per the result of problem 547 ( see link below)
https://gogeometry.blogspot.com/2010/12/problem-547-triangle-transversal.html
These 4 circles will meet at a point G
So 4 red circles will meet at a point M , image of G in this geometric inversion
Such that PM.PG= k