MathDB

6

Part of 2015 Oral Moscow Geometry Olympiad

Problems(2)

prove that 2 points are the feet of the altitudes

Source: Oral Moscow Geometry Olympiad 2015 grades 8-9 p6

8/8/2019
In the acute-angled non-isosceles triangle ABCABC, the height AHAH is drawn. Points B1B_1 and C1C_1 are marked on the sides ACAC and ABAB, respectively, so that HAHA is the angle bisector of B1HC1B_1HC_1 and quadrangle BC1B1CBC_1B_1C is cyclic. Prove that B1B_1 and C1C_1 are feet of the altitudes of triangle ABCABC.
geometryangle bisectoraltitudescyclic quadrilateral
3 circumcircles and a circle of given diameter concurrent, altitudes related

Source: Oral Moscow Geometry Olympiad 2015 grades 10-11 p6

8/6/2019
In an acute-angled isosceles triangle ABCABC, altitudes CC1CC_1 and BB1BB_1 intersect the line passing through the vertex AA and parallel to the line BCBC, at points PP and QQ. Let A0A_0 be the midpoint of side BCBC, and AA1AA_1 the altitude. Lines A0C1A_0C_1 and A0B1A_0B_1 intersect line PQPQ at points KK and LL. Prove that the circles circumscribed around triangles PQA1,KLA0,A1B1C1PQA_1, KLA_0, A_1B_1C_1 and a circle with a diameter AA1AA_1 intersect at one point.
isoscelesgeometryconcurrencyconcurrentcircumcirclecirclesaltitudes