For $\Delta ABC$ right-angled triangle at $A$ have $AB=AC$. Calculate the ratio of $MA: MB: MC$

Let $\,\widehat{MBA}=\alpha=\widehat{MCB}=\widehat{MAC}\,$, then:

$$\require{cancel} \widehat{AMB}= \pi - \widehat{MBA}-\widehat{MAB}= \pi - \widehat{MBA}-\left(\frac{\pi}{2}-\widehat{MAC}\right)=\pi-\bcancel{\alpha}-\left(\frac{\pi}{2}-\bcancel{\alpha}\right)=\frac{\pi}{2} $$

Similarly, $\,\widehat{BMC}=\pi-\left(\cfrac{\pi}{4}-\bcancel{\alpha}\right)-\bcancel{\alpha}=\cfrac{3\pi}{4}\,$, so $\,\widehat{CMA}=2 \pi - \widehat{AMB} - \widehat{BMC}=\cfrac{3\pi}{4}\,$.

Then, by the law of sines:

• $\;\triangle MAB\,$: $\;\;MA / \sin(\alpha) = AB / \sin(\pi/2) = AB$

• $\;\triangle MBC\,$: $\;\;MB / \sin(\alpha) = BC / \sin(3\pi/4) = \sqrt{2} \,BC = 2 \,AB$

• $\;\triangle MCA\,$: $\;\;MC / \sin(\alpha) = AC / \sin(3\pi/4) = \sqrt{2} \,AC = \sqrt{2} \,AB$

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[ EDIT  (+fix) ]  May also be worth noting that the above gives $\alpha=\arctan(1/2)$, so the point $\,M\,$ is uniquely determined by the conditions, and the construction can be fully solved. Asking for just the ratios of the distances to the vertices looks more like a diversion.

Forget the circles.

Scale triangle $ABC$ so that $AB = 1,\;AC = 1,\;BC=\sqrt{2}$.

Let $\;u = MA,\;v = MB,\;w = MC$.

Let $\theta$ be the common degree measure of angles $MBA,\;MAC,\;MCB$.

Chasing some angles, we get . . . $$\angle MAC = \theta,\;\angle ACM = 45-\theta,\;\angle CMA=135$$ $$\angle MCB = \theta,\;\angle CBM = 45-\theta,\;\angle BMC=135$$ Thus, triangles $MCB$ and $MAC$ are similar, hence, equating ratios of corresponding sides, we get $$\frac{\sqrt{2}}{1}=\frac{v}{w}=\frac{w}{u}$$ It follows that $$u\;\colon v\;\colon w = 1\;\colon 2\;\colon \sqrt{2}$$