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
Parting shot by McWorter)
I thought it was PA that was equal and orthogonal to AQ (PAQ an isosceles right triangle).
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Posted by Charlie
on 2005-10-26 14:55:59 |