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On the same line (Posted on 2013-12-22 )

Let ABC be a triangle. Let the point P lie in the plane of ABC.
Let X, Y, Z be the feet of the perpendiculars from P to the lines BC, AC, AB respectively.
PROVE: : The points X, Y , Z are collinear if and only if P lies on the circumcircle of ABC .

Solution
| Comment 1 of 3

W.l.o.g., take A, B, C, P, to be the order of points on the circumcircle. AZPY is a cyclic quad. (since / AZP = / AYP = 90^{o} ) Therefore / PYZ = / PAZ (subtended by PZ) = / PCX (ABCP cyclic) (1) PCXY is a cyclic quad. (since / PYC = / PXC) Therefore / PYX + / PCX = 180^{o} (opposite angles) giving / PYX + / PYZ = 180^{o} (using (1)) So XYZ is a straight line (the good old Simson Line).Converse : If XYZ is a straight line, / PYZ = / PAZ (since AZPY cyclic) Also, / PYZ = / PCB (since PCXY cyclic) Therefore / PAZ = / PCB So A, B, C and P are concyclic. (exterior and opposite angles equal). QED
Posted by Harry
on 2013-12-23 14:28:05

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