Dividing a paper into 3 parts using folding

Question

So I saw this video : https://www.youtube.com/watch?v=V7UJxeMlowQ, and it got me wondering: Why does that method divide the paper into exactly 3 equal rectangles, so I turned it into a math problem:

Draw a line from one vertex of a square to the midpoint of the opposite side. Draw a perpendicular bisector of the line you've drawn previously. Prove that the intersection of side of the square reflected over the perpendicular bisector to the other side of the square divides the side of the square it intersects in ratio $1:2$

Here is the picture I drew in Geogebra, in case the wording isn't clear:

The line $BC$ is reflected over the perpendicular bisector of $EC$, and the line $EI$ is formed. Prove that $AI$=$2IB$

Please don't use trigonometry while solving this problem, I'm not particulary good at it and I won't understand it.

Answer 1

If you denote by $M$ the centre of $EC$, perpendicular meets $BC$ in $N$, and $CD$ in $K$. By Pythagorean theorem, $EC=\frac{a\sqrt{5}}{2}$, and $MC=\frac{a\sqrt{5}}{4}$. By symmetry $\triangle EDC\sim\triangle KMC$, we have $\frac{KM}{MC}= \frac{ED}{DC}= \frac{1}{2}$, so $KM= \frac{a\sqrt{5}}{8}$. Then $KC= \frac{5a}{8}$ by Pythagorean theorem. By symmetry $\triangle KCN\sim\triangle EDC$ we have $\frac{CN}{CK}= \frac{DC}{DE}=2$, so $CN=\frac{5a}{4}$. Hence, $BN=CN-a=\frac{a}{4}$. By symmetry $\triangle IBN\sim \triangle IAE$ we finally have $\frac{IB}{IA}= \frac{BN}{AE}= \frac{1}{2}$.

Answer 2

If you like analytic geometry, you can set the coordinate system: $B(0,0)$, $A(0,a)$, $C(a,0)$. Then $E(\frac{a}{2},a)$, and $EC:\ y=-2x+2a$. If $M$ is a centre of $EC$, then $M(\frac{3a}{4},\frac{a}{2})$, and perpendicular through $M$ is $p:\ y=\frac{1}{2}x+\frac{a}{8}$. $p$ intersects $BC$ in $N(-\frac{a}{4},0)$, and $EN:\ y=\frac{4}{3}x+\frac{a}{3}$. The intersection of $EN$ and $AB$ is $I(0,\frac{a}{3})$, which implies $AI=2IB$.