In the diagram, OA is perpendicular to PQ. O is the centre of the circle. AC and EP meet at B on the circle and QC and AE meet at D on the circle. The goal is to prove that AP=AQ.
I had tried to prove this by drawing a line through E parallel to PQ and I had proved some of the triangles were similar. However, this is the part where I got stuck.
I drew a line parallel to PQ through E. I named the meeting point between PE and QC "G", between the line produced at E and CQ "F" and between CA and the line produced at E "H". I then proved that triangle CBG is similar to DEG, CAD is similar to BAE, HBE is similar to ABP, FDE is similar to ADQ and FGE is similar to PGQ. It is at this point when I got stuck.
Note: the points are renamed in figure.
You can use this property that in a triangle the bisector of an angle meets the perpendicular bisect of the side opposite to that on circumcircle. As shown in figure, the line connecting D to B crosses th circle at I and a perpendicular from I to line g(perpendicular on AB) meets circle at E. The bisectors of angles EFI of triangle EFI and EGI of triangle EGI meet at J on the circle. We have:
1):$\angle LFH=\angle LGH$
2):$\angle HFI=\angle HGI$
Also GB and extension of DI meet at B, the construction of triangle DEI is symmetric about DC.
3):$\angle EGL=\angle EFL$
From relation 3 and the relation and fact (2) we may conclude that A must be on extension of DE which means CD is perpendicular bisector of AB and C is mid point of AB.