Regarding a geometrical proof of irrationality of $\sqrt 2$ based on $\gcd$.

Question

In the below image stated on article's page 17, on $\gcd$ there is proof for irrationality of $\sqrt 2$ here. Have some questions:
1. I am unable to understand how it is deduced geometrically that $|EB| = |EF|$.
2. How it is deduced that $\gcd(AC,AB)=\gcd(CE,CF).$
3. How it is deduced that $|EC|\gt |EB|$, or is it just an hypothesis/assumption.

Answer 1

1. It's assumed $AB=AF$. Since angles at $B$ and $F$ are also equal and the the triangles $ABE$ and $AFE$ share the $EF$ length, their third sides are equal, too, by Pythagorean theorem. Hence the two triangles are congruent.

2. $AC=AF+CF=AB+CF$ – and if a sum ($AC$) shares a divisor with one addend ($AB$) then it must share the same divisor with another one ($CF$).
   On the other hand $AB=CB=CE+BE=CE+CF$, so similarly the common divisor of a sum and one addend must be a divisor of another addend, so $\gcd(AB,CF)=\gcd(CE,CF).$
   Hence $\gcd(AC,AB)=\gcd(AB,CF)=\gcd(CE,CF) $

3. $EB=EF$ and $EF$ is a side of a small square, while $EC$ is its diagonal, and a diagonal is longer than the side.

Answer 2

The triangle $ABF$ is isosceles on the basis $BF$ by construction; therefore the angles $\widehat{AFB}$ and $\widehat{ABF}$ are equal.

The angles $\widehat{ABE}$ and $\widehat{AFE}$ are right by construction. Thus $\widehat{EFB}$ and $\widehat{EBF}$ are equal because complementary to equal angles. Hence $EFB$ is an isosceles triangle on the basis $BF$; therefore $EF=EB$.

Your second question stems from the fact that, if $a>b$, then $\gcd(a,b)=\gcd(a-b,b)$ (the very fact on which the Euclidean algorithm is based).

Finally, $EC>EF$ because the diagonal in a square is greater than the side; and we proved $EF=EB$, so $EC>EB$.