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 re(2): Parting shot by McWorter)

I see from trying in Geometer's Sketchpad that in fact those triangles are congruent.  But I think it takes some steps of proof to show that those sides that are diagonals of the parallelograms are in fact equal to the sides of the original triangle that are not equal to the respective sides of the parallelograms.

Also, the orthogonality needs some steps of proof. 

I think after these are added, the whole proof could be as lengthy as the ones already offered.

Posted by Charlie on 2005-10-27 14:04:30