Let $A_{2}E \bot AC$ We have $\triangle A_{2}EA=\triangle ACB$ so $A_{2}E/EC=A_{2}E/(EA+AC)=AC/(AC+BC)$ Otherwise $FC/CC_{1}=FC/BC=AC/(AC+BB_{2})=AC/(AC+BC)$ So $A_{2}E/EC=FC/CC_{1}$ Thus triangle $A_{2}EC$ and $FCC_{1}$ are similar So that $\angle A_{2}CE=\angle FC_{1}C$. Thus $A_{2}C//FC_{1}$ Let $AD$ be the bisector of angle $ACB$ $AD/DB=AC/CB=AC/CC_{1}=AM/MF$ so $DM//BC$ $(1)$ $AD/DB=AC/CB=AC/BB_{2}=CF/FB$ so $DF//AC$ $(2)$ Form (2) $CF/CB=AD/AB=DF/BB_{2}$ so $CF=DF$ $(3)$ From (1) (2) (3) and $\angle ACB=90$ we have $CNDF$ is a square $q.e.d$

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I labelled randomly. Let $ b=CB$ and $ a=AC$. First, note by homothety that NM, BC, and AE are concurrent. Now, we prove that AE ML CH are concurrent, by clarifying O to be the intersection of ONLY AE and CH, and declaring L as the foot of the perpenicular from O. By find the length LC, we can then prove that this is the same as the side length of the square. First: OL/AL = DE/DA = $ \frac{b}{a+b}$. Next: OL/LC = JA/(AC+JH) = $ \frac{a}{a+b}$ Thus AL/LC = a/b = AC/BC = AL/LN $ \implies \boxed{LC=LN}$ QED

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