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 Isosceles Right Triangles (Posted on 2005-10-19)
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.

 See The Solution Submitted by Bractals Rating: 3.2500 (4 votes)

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 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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