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(3): Parting shot by Charlie)

You remind me of intelligent design theorists trying to prove the existence of God.  According to this theory, if you cannot figure out how something was created, then it must have been created by a higher power.  Since you cannot find an easy explanation for Bractal's problem, you conclude (admittedly tentatively) that the problem must have a more complex, or "higher", solution.

Judge for yourself.  Is the proof below, of what Geometer's Sketchpad confirmed, as "lengthy", or as complex, as "the ones already offered"?

Slide triangle ABC along BL until B coincides with L.  Then rotate ABC 90 degrees; whence, by how the figure is constructed, A will coincide with P and C will coincide with (the old) B.  Thus triangle PLB is ABC rotated 90 degrees and so PB is equal and orthogonal to AC.  Similarly, triangle CQN is congruent to ABC and CQ is equal and orthogonal to AB.

Posted by McWorter on 2005-10-28 19:56:29