[tex]m( \angle CBD)= m(\angle DCB)=22 \textdegree 30' \Rightarrow \DeltaDBC-isoscel \Rightarrow BD=CD. \\ \\ m( \angle BDC)=180 \textdegree-(m( \angle DBC)+m( \angle BCD))=180 \textdegree -45 \textdegree =135 \textdegree. \\ \\ m( \angle BDA)=180 \textdegree- m( \angle BDC)=180 \textdegree -135 \textdegree =45 \textdegree \Rightarrow \\ \\ \Rightarrow \Delta ABD-dreptunghic~isoscel \Rightarrow AD=AB~si~BD=AB \sqrt{2} \Leftrightarrow \\ \\ \Leftrightarrow CD=AB \sqrt{2}.[/tex]
[tex]AC-AB=AC-AD=CD=AB \sqrt{2}. \\ \\ AC+AB=AD+CD+AB=AB+AB \sqrt{2}+AB=AB(2+ \sqrt{2}). \\ \\ \frac{AC-AB}{AC+AB}= \frac{AB \sqrt{2}}{AB(2+ \sqrt{2})} = \frac{ \sqrt{2}}{2+ \sqrt{2}}= \frac{ \sqrt{2}(2- \sqrt{2})}{(2+ \sqrt{2})(2- \sqrt{2})}= \frac{2 \sqrt{2}-2}{4-2} = \frac{2( \sqrt{2}-1)}{2}= \\ \\ =\sqrt{2}-1. [/tex]
