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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.
| 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 = 90o)
Therefore /PYZ = /PAZ (subtended by PZ)
= /PCX (ABCP cyclic) (1)
PCXY is a cyclic quad. (since /PYC = /PXC)
Therefore /PYX + /PCX = 180o (opposite angles)
giving /PYX + /PYZ = 180o (using (1))
So XYZ is a straight line (the good old Simson Line).
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).
Posted by Harry
on 2013-12-23 14:28:05
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