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.

(In reply to Analytic solution by Jer)

It's the coordinates of B (not C) that you're calling (x,y).

It's M (not D) that's at (0,-1).

Posted by Charlie on 2005-10-19 14:02:19