Euclid, Schmeuclid

euclid schmeuclid fallacy

Consider any two circles intersecting at points Q and R. Their diameters are QP and QS. PS intersects the circles at M and N.

Now, any angle inscribed in a semicircle is a right angle. This means that both QNP and QMS are right angles — and thus that there are two perpendiculars from Q to PS. Doesn’t it?