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Amazing (Posted on 2013-12-07) Difficulty: 4 of 5

  
Let C and D be distinct points on the semicircle with diameter AB.
Let rays AC and BD intersect at point E and let F be the foot of the
the perpendicular from E to AB. Let P be the intersection of the
perpendicular bisector of AB and line CD and Q a point on CD,
distinct from P.

Prove that FQ ⊥ CD if and only if ∠PAQ = ∠PBQ.
  

  Submitted by Bractals    
Rating: 3.0000 (1 votes)
Solution: (Hide)

  
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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  Subject Author Date
A specific caseJer2013-12-10 16:19:51
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