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.
  

An Algebra solution is getting very messy so I tried it with some actual numbers:
The circle = the unit circle: y = √1-x²
A=(1,0)
B=(-1,0)
C=(-4/5,3/5)
D=(5/13,12/13)
Gives
P=(0,9/11)
E=(-1/3,4/9)
F=(-1/3,0)
Setting Q perpendicular gives
Q=(-101/195,44/65)
With a little vector work ∠PAQ and ∠PBQ both come out to
arccos(11√(1170)/390) ≈ 15.255º

I'll try again with C=(xc,yc) and D=(xd,yd) soon if I get a chance.

Posted by Jer on 2013-12-10 16:19:51