Xét tam giác NIO có $OI=NI.\cos \beta =\frac{a}{2}\cos \beta ,NO=NI.\sin \beta =\frac{a}{2}\sin \beta$

Xét tam giác SEF vuông tại S có

$$\begin{gathered} \widehat{SEF}=\widehat{ESM}+\widehat{SME}=45°+90°-\beta =135°-\beta \\ SF=SE.\tan \widehat{SEF}=SE.\tan \left(135°-\beta \right)=SE.\frac{1+\tan \beta }{\tan \beta -1} \end{gathered}$$

Vì SI là độ dài đường phân giác trong góc $\widehat{FSE}$ nên

$$\begin{gathered} SI=\sqrt{2}.\frac{SE.SF}{SE+SF}\iff \frac{a}{2}\cos \beta =\sqrt{2}\frac{SE\tan \left(135°-\beta \right)}{1+\tan \left(135°-\beta \right)} \\ \iff SE=\frac{a\left[1+\frac{1+\tan \beta }{\tan \beta -1}\right]\cos \beta }{2\sqrt{2}\frac{1+\tan \beta }{\tan \beta -1}}=\frac{a\sin \beta }{\sqrt{2}\left(1+\tan \beta \right)} \end{gathered}$$

Do đó $EF=\frac{SE}{\cos \widehat{SEF}}=\frac{SE}{\cos \left(135°-\beta \right)}=\frac{a\sin \beta }{\left(1+\tan \beta \right)\left(-\cos \beta +\sin \beta \right)}=-\frac{a}{2}\tan 2\beta$