DEFINITION

If X is a point not on line YZ, then
the foot of the perpendicular from X
to YZ is the point X' on YZ such that
XX' ⊥ YZ.

PROVE THE FOLLOWING

If A, B, C, and D are distinct points
on a circle such that lines AB and CD
are not perpendicular, then the feet
of the perpendiculars from A and B to
to CD and the feet of the perpendiculars
from C and D to AB lie on a circle.

Let  A’, B’, C’, D’, be the feet of the perpendiculars from A, B, C and D.

If AB and CD are parallel, the quadrilateral A’B’C’D’ is a reflection of ABCD and therefore also cyclic.

Otherwise, let P be the point of intersection of AB and CD. The right-angled triangles PAA’, PBB’, PCC’, PDD’ are similar (common angle at P), so:

|PA|/|PA’| = |PB|/|PB’| = |PC|/|PC’| = |PD|/|PD’|.

It follows that     (|PA||PB|)/(|PA’||PB’|)  =  (|PC||PD|)/(|PC’||PD’|)
                                                                       
Now, |PA||PB| = |PC||PD| (intersecting chord/secant theorem).

Therefore:  |PA’||PB’| = |PC’||PD’| and, using the converse of that theorem, it follows that A’B’C’D’ are concyclic points.

Posted by Harry on 2012-05-02 13:29:30