It is easy to divide a square into four congruent isosceles triangles, just draw both diagonals. But can you divide a square into four incongruent isosceles triangles?

(In reply to using geometer's sketchpad by Charlie)

Most of the construction is the same and the final picture is the same, but the proof is easier with a change to the final construction:

From the top left going clockwise, label the corners of the square A,B,C,D.

Draw a diagonal from A to C. The first isosceles triangle is in the upper right-- triangle ABC.

From A, lay off along the diagonal a length equal to one side of the square (call it length 1, for a unit square), and label the point E. The second isosceles triangle is AED.

The remaining length of the diagonal, EC, is sqrt(2) - 1. 

Bisect angle DAC, and label the intersection of the bisector with CD, F. And DAC is 45 degrees, with sine = 1/sqrt(2) and cosine = 1/sqrt(2). By half-angle formula tan(x/2) = (1 - cos x) / sin x. This comes out to sqrt(2) - 1.  By the symmetry of AEFD (which can be proved by making an additional point where the bisector cuts ED and showing congruent triangles), EF is also sqrt(2) -1. That shows that both triangles ECF and EFD are isosceles.

Edited on March 23, 2007, 3:10 pm

Posted by Charlie on 2007-03-23 15:06:34