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Geometry Problem 989: Arbelos, Semicircles, Diameter, Perpendicular, 90 Degree, Common External Tangent, Rectangle, Midpoint of Arc.

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The figure shows an arbelos ABC (AB, BC, and AC are semicircles), BD is perpendicular to AC and chord EF is the common external tangent to semicircles AB and BC at G and H. Prove that: (1) BGDH is a rectangle; (2) D is the midpoint of arc EF.
 

Geometry Problem 989: Arbelos, Semicircles, Diameter, Perpendicular, Common External Tangent, Rectangle, Midpoint of Arc
 

Home | SearchGeometry | Problems | All Problems | Open Problems | Visual Index | 10 Problems | Problems Art GalleryArt | 981-990 | Arbelos | Semicircle | Perpendicular lines | Tangent Circles | Circle Tangent Line | Rectangle | Midpoint | Email | Solution / comment | by Antonio Gutierrez
Last updated: Dec 27, 2014