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15.3

Part of Geometry Mathley 2011-12

Problems(1)

Geometry Mathley 15.3 OH^2/ R^2=OM/ OD +ON /OE + OP / OF

Source:

6/14/2020
Triangle ABCABC has circumcircle (O,R)(O,R), and orthocenter HH. The symmedians through A,B,CA,B,C meet the perpendicular bisectors of BC,CA,ABBC,CA,AB at D,E,FD,E, F respectively. Let M,N,PM,N, P be the perpendicular projections of H on the line OD,OE,OF.OD,OE,OF. Prove that OH2R2=OM‾OD‾+ON‾OE‾+OP‾OF‾\frac{OH^2}{R^2} =\frac{\overline{OM}}{\overline{OD}}+\frac{\overline{ON}}{\overline{OE}} +\frac{\overline{OP}}{\overline{OF}} Đỗ Thanh Sơn
geometrycircumcirclesymmedianperpendicular bisector