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Geometry Problem 1150: Circle, Perpendicular Secants, 90 Degrees, Congruence. Level: School, College, Mathematics Education

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The figure below shows a circle O and the perpendicular secants ABC and ADE. DF is perpendicular to BE, CG perpendicular to EB extended, and AH perpendicular to CD. Prove that GF = 2AH.
 
 

Geometry Problem 1150: Circle, Perpendicular Secants, 90 Degrees, Congruence



 

See also:
Isolines of Problem 1150
GeoGebra of problem 1150
 

Geometric Art: Hyperbolic Kaleidoscope of problem 1050

Hyperbolic Kaleidoscope of problem 1150

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