Angle addition

See figure. Construct 2 squares, EHGF and EIJF. We know that A=45 degrees. We let HJG=e degrees. and we also know that e+C=45 degrees= A We know that OH/OB=GH/GJ=1/2, thus angle B = angle e. Therefore, B=e and we can conclude that C+e=C+B=45 degrees, and A+B+C=90 degrees.

As shown above, in order to solve the problem, we must first double the figure, and construct a triangle QGD. In order to continue, we must first prove QGD is a right isosceles, thus proving the angle d is right and both others equate to 45°.

We do so by observing that the triangles IQG, GFD and AQC are all equal. Their legs are 2L and L, where L equals to the side of one of the squares. Thus, their hypothenuses are equal too. The sides GD and QD are the hypotenuses of two of the former triangles, and thus are equal.

In order to prove angle d is right, we must also notice that since angle b is contained in the triangle AQC, it is thus contained in its mirrored versions at QIG and GFD. The angle e, is the same angle as b. In the 90° angle CDE, we must also plug in b. Now, we have that b+c+45°=90°, if that third angle really is 45°

By observing that b+f+90°=180° at GFD, we conclude that f=b-90°. And if, obviously, d+e(which is b)+f=180°, we conclude d=90°, thus proving the other angles of QGD are equal to 45° and proving that the answer is 90°

Of course one can easily solve this problem, and in a much faster way, with trigonometry, but this is a rather nice and creative alternative.