$\angle ABE= \angle CBF$ and $\angle ABC= 90 \implies \angle EBF=90$ $\triangle ABE \equiv \triangle CBF \implies BE=BF \implies \angle BEF= \angle BFE =45 \implies \angle AEB= \angle BFC=135$ Let it be $\angle BAE= \angle BCF=x$ and $\angle ABE=\angle CBF=45-x$ By the law of sines in $\triangle ABE \frac{AB}{sin 135} = \frac{EB}{sin x}$ Analougsly in $\triangle BEF \frac{EB}{sin 45} = \frac{EF}{sin 90} \iff EB=EF \cdot sin 45$ $\frac{AB}{sin 180-135=45}=\frac{EF \cdot sin 45}{sin x}$ But $AB= EF \implies \frac{1}{sin 45}=\frac{sin 45}{sin x}$ $\implies sin x= sin(45)^2= \frac{\sqrt{2}^2}{4}=\frac{2}{4}=\frac{1}{2}=sin 30$ $\implies x=30$ and $45-x=15$