Two unequal circles whose centers are O and T intersect at A and B. S is the midpoint of their line of centers. A line drawn thru A perpendicular to SA meets the circles at P and Q respectively. Prove that PA =PQ

Let OX be a line through O, parallel to SA, meeting PA at X. Then OX is perpendicular to PA and it follows that X is the midpoint of PA.

Similarly, let TY be a line through T, parallel to SA, meeting AQ at Y. Then TY is perpendicular to PA and it follows that Y is the midpoint of AQ.

Now you should be able to see that, since S is the midpoint of OT, it follows that A is the midpoint of XY. Thus PX=XA=AY=YQ, and so PA=AQ.

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