Question 480043
<pre>

Every isosceles right triangle, a.k.a, a 45-45-90 right triangle, is half 
a square cut down a diagonal.

Draw a square, which has 4 interior 90<sup>o</sup> angles.

{{{drawing(150,150,-1.2,1.2,-1.2,1.2,

line(-1,-1,-1,1), line(-1,1,1,1),line(1,1,1,-1),line(1,-1,-1,-1),

locate(.6,-.6,90^o),locate(.6,1.02,90^o), 

locate(-.95,-.6,90^o),locate(-.95,1.02,90^o)


)}}} 

Cut it down one of the diagonals:

{{{drawing(150,150,-1.2,1.2,-1.2,1.2,

line(-1,-1,-1,1), line(-1,1,1,1),line(1,1,1,-1),line(1,-1,-1,-1),

line(-1,1,1,-1), 

locate(.6,-.6,90^o),locate(.6,1.02,90^o), 

locate(-.95,-.6,90^o),locate(-.95,1.02,90^o)


)}}}

Take the left half of the square:

{{{drawing(150,150,-1.2,1.2,-1.2,1.2,

line(-1,-1,-1,1),line(-1,1,1,-1),

line(-1,-1,1,-1), 

locate(.3,-.6,45^o), 

locate(-.95,-.6,90^o),locate(-.95,.7,45^o)


)}}}

The triangle above is an isosceles right triangle.  Notice that its
vertex angle is 90<sup>o</sup> and each base angle is 45<sup>o</sup>.
As you see, the base angles CANNOT be {{{cross(60^o)}}} as you have stated.

Edwin</pre>