Construct a square ABCD as shown. Draw the diagonal AC. Now if AC and AB are commensurable, then there is a line segment of some length, say r, such that AC=m1r and AB=m2r where m1 and m2 are integers. Now draw a circle centered at A with radius AB. Note that CB and CD are tangent to this circle [why?]. This circle intersects AC at some point; call it P. Now obviously AP=AB, so PC=AC-AB= m1r - m2r= (m1 - m2)r so PC is also an integer multiple of r and is therefore commensurable with AB and AC. Construct the tangent to the circle at P. Note that this tangent is perpendicular to AC. This tangent intersects BC at some point; call it Q. Now Angle(PCQ)=45o=Angle(PQC) so Triangle(PQC) is an isosceles triangle and therefore PC=PQ. We can construct a square with vertices C, P, and Q; call the fourth vertex R (see the figure above). Now since QP and QB are tangents to a circle from an external point, they are equal, so QB=QP=PC=(m1 - m2)r=m4r where we have set m4= m1 - m2. Then QC=BC-QB=m2r - m4r=(m2 - m4)r=m3r (with m3=m2 - m4) so QC is also commensurable with AB and AC. We now have a new square whose side and diagonal are commensurable with AB and AC. We note that m1>m2>m3>m4. There's no reason we can't do the same thing on square PCRQ. We can draw a circle centered at C with radius CP, call the intersection of this circle with CQ point X, construct the tangent there, and construct square XYZQ as shown. Then the side and diagonal of this square are commensurable with AB and AC as well. Then we can continue the process indefinitely constructing smaller and smaller squares with sides and diagonals commensurable with AB and AC. However, there are only a finite number of integer multiples of r less than m1r, so we have a contradiction.