Let BAC be an arbitrary triangle with external squares ABIJ, BCKL, and CAMN. IBLP, KCNQ, and MAJR are parallelograms. Prove that PAQ, QBR, and RCP are isosceles right triangles.
Let's call x=B-A; y=C-B, and z=A-C. It holds that x+y+z=0. Let's also write v' for vector v rotated by 90 degrees.
Then, we have I=B+x', J=A+x', K=C+y', L=B+y', M=A+z', and N=C+z'.
Looking at the parallelograms, since L-B=P-I, we have P=L+I-B;
similarly, Q=K+N-C, and R=J+M-A.
We want to prove that C-R is equal and ortogonal to P-C. First, C-R=
C-J-M+A= C-A-x'-A-z'+A= C-A+y'= -z+y'. Then, P-C= L+I-B-C=
B+y'+B+x'-B-C= B-C-z'= -y-z'.
Now, rotating -z produces -z', and rotating y' produces -y, so... QED.
PS. By the way, out of the intermediate results we can find other
interesting "theorems". For example, A-R=y', so AR equals BC, and is
ortogonal to it, and similar conditions apply to the other points.
Vectorial solution, plus new theorems
