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3

Part of 2009 Oral Moscow Geometry Olympiad

Problems(2)

<AA_1K=<BB_1M wanted, B_1K// BC, A_1M// AC, altitudes AA_1,BB_1

Source: 2009 Oral Moscow Geometry Olympiad grades 8-9 p3

9/19/2020
In the triangle ABCABC, AA1AA_1 and BB1BB_1 are altitudes. On the side ABAB , points MM and KK are selected so that B1K∥BCB_1K \parallel BC and A1M∥ACA_1M \parallel AC. Prove that the angle AA1KAA_1K is equal to the angle BB1MBB_1M.
(D. Prokopenko)
geometryequal anglesparallelatltitudes
incenter lies on perpendicular, other incircle related

Source: 2009 Oral Moscow Geometry Olympiad grades 10-11 p3

9/14/2020
Altitudes AA1AA_1 and BB1BB_1 are drawn in the acute-angled triangle ABCABC. Prove that the perpendicular drawn from the touchpoint of the inscribed circle with the side BCBC, on the line ACAC passes through the center of the inscribed circle of the triangle A1CB1A_1CB_1.
(V. Protasov)
geometryincenterperpendicularincircle