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Isosceles Right Triangles (Posted on 2005-10-19) Difficulty: 3 of 5
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.

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Solution Vectorial solution, plus new theorems | Comment 3 of 10 |
Let's call x=B-A; y=C-B, and z=A-C. It holds that x+y+z=0. Let's also write v' for vector v rotated by 90 degrees.

Then, we have I=B+x', J=A+x', K=C+y', L=B+y', M=A+z', and N=C+z'. Looking at the parallelograms, since L-B=P-I, we have P=L+I-B; similarly, Q=K+N-C, and R=J+M-A.

We want to prove that C-R is equal and ortogonal to P-C. First, C-R= C-J-M+A= C-A-x'-A-z'+A= C-A+y'= -z+y'. Then, P-C= L+I-B-C= B+y'+B+x'-B-C= B-C-z'= -y-z'.

Now, rotating -z produces -z', and rotating y'  produces -y, so... QED.

PS. By the way, out of the intermediate results we can find other interesting "theorems". For example, A-R=y', so AR equals BC, and is ortogonal to it, and similar conditions apply to the other points.

  Posted by Federico Kereki on 2005-10-19 19:59:27
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