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.

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