On the sides $BC$ and $AB$ of the triangle $ABC$ the points ${{A} _ {1}}$ and ${{C} _ {1}} $ are selected accordingly so that the segments $A {{A} _ {1}}$ and $C {{C} _ {1}}$ are equal and perpendicular. Prove that if $\angle ABC = 45 {} ^ \circ$, then $AC = A {{A} _ {1}} $. (Gogolev Andrew)