Euclid 1999 Question 4(a) - Circle Tangent Intersection Below is a question and the intended solution to a math contest problem.

I understand that if for both circles, if you assume that a circle's centre, the two points on the circumference that touch a tangent line each, and the intersection of the tangent lines; that these four points form a square, then the distances can be calculated trivially.
Thus, I would like to know how it is known that the points form two squares.
To clarify, this is what I would like to know:

Given a circle and two perpendicular lines both tangent to the circle, how is it known that the tangent points + center + intersection point forms a square?


You are given that the angle where tangents intersect is $90^\circ$.  Concentrate on the larger circle, specifically where the upper tangent contacts it.  By the definition of a tangent you know that the tangent is at right angles to radius drawn there.  So that is another $90^\circ$.  The same argument applies at the contact of the lower point.  That's $270^\circ$ total.  Now, the intersection of the two tangents together with the two contact points and the circle center form a quadrilateral.  The interior angles of any quadrilateral add to $360^\circ$.  We have accounted for $270^\circ$, leaving $90^\circ$.  So we have a rectangle, at least.  But the two radii are equal in length.  So we have a square.
