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.

Without loss of generality, scale the triangle so that AC=1 and place A at (0,0), C at (1,0) and B in the first quadrant.

Call the coordinates of C (x,y)

Then we have the following:

N=(1,-1)
D=(0,-1)
R=(-y,x-1)
Q=(y+1,-x)
etc..

The distances can easily be checked as equal
BQ = BR = sqrt((-x+y+1)^2 + (x+y)^2

The slopes of segments BR and BQ are
(-x+y+1)/(x+y) and (x+y)/(x-y-1)
The product of these being -1

Therefore Triangle QBR is an isosceles right triangle with right angle B

Posted by Jer on 2005-10-19 13:36:12