Shrink

The solution to this problem only uses two very simple facts:

1. The diagonals of a rectangle are equal in length.
2. For any triangle, the shortest distance from a vertex to the opposite side is the perpendicular to that side.

Putting the two together, for any right angled triangle the length of P Q is a minimum when X is the foot of the perpendicular from A to B C. As the point X moves along the hypotenuse the rectangle A P X Q changes but P Q is always equal in length to A X.

As a further challenge, you may like to prove that the position of X is given by: ( C X/ X B) = ( C A/ A B)².