Let AHA, BHB, CHC be the altitudes of triangle ABC. The extensions of AHA, BHB, and CHC intersect the circumcircle O at A1, B1, and C1, respectively. Prove that (1) HAHB // A1B1, HBHC // B1C1, and HAHC // A1C1; (2) the area of triangle A1B1C1 is 4 times the area of triangle HAHBHC.
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Ten problems: 1411-1420
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HTML5 and Dynamic Geometry
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