If CD || AB, then F = O and P = Q. This would contradict the proviso
that P ≠ Q. Therefore, let line CD intersect line AB at point H.
WLOG we can assume that A lies between H and B.
With the positive real numbers a, h, and p
[ a < h and p < a/√(1 - [a/h]2) ]
defining the coordinates of A(-a,0), B(a,0), H(-h,0), O(0,0), and P(p,0);
all other points of the problem are determined.
The line CD is defined by
y = p(x + h)/h (1)
If the coordinates (f,0) are assumed for point F, then the coordinate
qx for point Q (the foot of the perpendicular from F to line CD)
is determined by the following equations:
qy = p(qx + h)/h and (qy - 0)/(qx - f) = -h/p.
Giving qx = h(fh - p2)/(h2 + p2).
(2)
If ∠PAQ = ∠PBQ, the point Q must lie on the circumcircle of ΔABP.
Whose equation is
x2 + y2 + y(a2 - p2)/p = a2
(3)
Combining equations (1) and (3) we get
[(h2 + p2)x + h(a2 + p2)]x = 0.
The x = 0 corresponds to point P and the other to point Q,
qx = -h(a2 + p2)/(h2 + p2).
(4)
Therefore, if
the qx of (2) equals the qx of (4),
(5)
then
FQ ⊥ CD if and only if ∠PAQ = ∠PBQ
But, (5) is true only if f = -a2/h ( or F and H are inverse points
with respect to the circle with diameter AB). We will now prove this
to complete the proof.
The points F and H are inverse points with respect to the circle with
diameter AB if |OF||OH| = a2.
ANGLE EQUIVALENCES:
∠CDB = ∠CAB = ∠EAF
∠DCA = ∠DBA = ∠EBF
∠AHD = ∠CAB - ∠DCA = ∠EAF - ∠EBF
∠ADH = 180° - ∠ADE - ∠CDB = 90° - ∠EAF
FROM ΔABE:
(a - |OF|)tan(∠EAF)
= (|OA| - |OF|)tan(∠EAF)
= |AF|tan(∠EAF)
= |EF|
= |BF|tan(∠EBF)
= (|OB| + |OF|)tan(∠EBF)
= (a + |OF|)tan(∠EBF)
a[tan(∠EAF) - tan(∠EBF)]
∴ |OF| = ----------------------------
tan(∠EAF) + tan(∠EBF)
FROM ΔABD AND ΔADH
2a sin(∠EBF)/sin(∠EAF - ∠EBF)
= |AB| sin(∠ABD)/sin(∠AHD)
= |AD|/sin(∠AHD)
= |AH|/sin(∠ADH)
= (|OH| - |OA|)/sin(∠ADH)
= (|OH| - a)/sin(90° - ∠EAF)
= (|OH| - a)/cos(∠EAF)
a[tan(∠EAF) + tan(∠EBF)]
∴ |OH| = ----------------------------
tan(∠EAF) - tan(∠EBF)
Therefore, |OF||OH| = a2
QED
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