On ray BM, draw point N so AM=MN (1)

Because \(\widehat{BAM}+\widehat{MAC}=\widehat{BAC}=90^0\Rightarrow\widehat{BAM}+\widehat{MBA}=90^0\) (\(\widehat{MAC}=\widehat{MBA}\))

\(\Rightarrow\widehat{AMB}=90^0\Rightarrow\)\(\Delta AMB\) is a square triangle.

So \(\Delta MAN\) is an isosceles square triangle.

\(\Rightarrow\widehat{MAN}=\widehat{MNA}=45^0\Rightarrow\widehat{BAN}+\widehat{MAC}=45^0\)

But \(\widehat{ACM}+\widehat{MCB}=\widehat{ACB}=45^0\)

\(\Rightarrow\widehat{BAN}+\widehat{MAC}=\widehat{ACM}+\widehat{MCB}=45^0\)

\(\widehat{MAC}=\widehat{MCB}\Rightarrow\widehat{BAN}=\widehat{ACM}\)

\(\Rightarrow\Delta ABN=\Delta CAM\) (Angular angle) \(\Rightarrow BN=AM\) (2)

By (1) and (2) \(\Rightarrow AM=MN=BN\Rightarrow MB=2MA\)

So that \(MA:MB=1:2\) (*)

Consider \(\Delta MAN\): 

\(\widehat{AMN}=90^0\Rightarrow AM^2+MN^2=AN^2\Leftrightarrow2AM^2=AN^2\)(Pytagoras theorem)

\(\Rightarrow\sqrt{2.AM^2}=\sqrt{AN^2}\Rightarrow\sqrt{2}.\sqrt{AM^2}=AN\)

\(\Leftrightarrow\sqrt{2}.AM=AN\). I have \(AN=MC\) (\(\Delta ABN=\Delta CAM\))

So \(MC=\sqrt{2}.AM\) \(\Rightarrow MA:MC=1:\sqrt{2}\) (**)

From (*) and (**) \(\Rightarrow MA:MB:MC=1:2:\sqrt{2}\).