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.

If we consider A, B, C, as points on the complex plane, since a 90 degrees counterclockwise rotation is the same as multiplying by -i we get I = B-i(B-A) = iA+(1-i)B, and J = (1+I)A-iB, and similar formulas for the other points.

For a parallelogram, the sum of opposite vertices is a constant, so I+L=B+P, and we find formulas for P, Q, and R.

Finally, P-A = -i(Q-A), proving that P-A is equal to Q-A, and that they are perpendicular.

Posted by Old Original Oskar! on 2005-10-19 22:01:31