Proof: a repeat

Since the proof has not been answered correctly yet, I am bringing it up. Taken from a book by famous American mathematician. Translated from Russian to English.

“Obtuse angle (>90°) is sometimes equal to a right angle (90°)”

Assume the figure ABCD is a square. Divide the segment AB in half and through the point of division (E) create a line perpendicular to AB. This line will intersect the opposite side of the square (DC) at F. Such that DF = FC.

From the point C, create a segment CG equal to CB. Connect points A and G with a line and divide the segment AG in half by a point H. Then, from the point H create a line HK which is perpendicular to the segment AG.

Because the segments AB and AG are not parallel, the lines EF and HK are also not parallel. Therefore, they will intersect at some point (call it K). Let’s connect the point K with points D, A, G and C.

Triangles KAH and KGH are congruent (equal?), because they have a common side HK, because AH = HG, and because the angles at H are right angles (both equal to 90°). [Here we are invoking a well-known theorem that says that if two triangles’ sides and the angle between those two sides are equal, the triangles are congruent.] Therefore, KA = KG.

Triangles KDF and KCF are also congruent, because they have a common side FK, because DF = FC, and because the angles at the point F are right (both equal to 90°). Therefore, KD = KC, and angle KDC is equal to angle KCD.

Besides that, DA = CB = CG.

Therefore, the sides of the triangles KDA and KCG are equal. Therefore, angles KDA and KCG are equal. Let’s subtract from them the equal angles KDC and KCD. It is obvious that the results of the subtraction should also be equal; i.e., that angle GCD = angle ADC.

But, the angle GCD is an obtuse angle (>90°). While the angle ADC is a right angle (90°).

Therefore, sometimes, obtuse angle is equal to the right angle, which is what was required to prove and what is so far from the truth.