On
Let $O$ be the center of the circle, $r$ be the radius, $h=MT$ and $x=MN$. We have $$2r=AN=AM+MN=a+x$$ $$r^2=MT^2+OM^2=h^2+(ON-MN)^2=h^2+(r-x)^2$$ $$2rx=h^2+x^2$$ $$(r+b)^2=(ON+NC)^2=OC^2=OT^2+TC^2=r^2+MT^2+MC^2=r^2+h^2+(MN+NC)^2=r^2+h^2+(x+b)^2$$ $$2br=h^2+x^2+2bx=2rx-x^2+x^2+2bx=2xr+2bx$$ $$2r(b-x)=2bx$$ $$(a+x)(b-x)=2bx$$ $$ab-bx-ax-x^2=0$$ $$x^2+x(a+b)-ab=0$$
$$x=\frac{1}{2}\left(\sqrt{a^2+6ab+b^2}-a-b\right)$$
Let $\xi=\angle BTL=\angle MTN$ and $\alpha=\angle BAC=\angle MBC=\angle TBL$. Then
$$(a+x+b)\cos^2\alpha=AC\cos^2\alpha=AM=a$$ $$\tan\xi=\frac{x}{h}$$ $$BM^2=AM\cdot MC=a(x+b)$$
By sine law in triangle $\triangle BLT$:
$$\frac{BL}{\sin\xi}=\frac{BT}{\sin(180-\alpha-\xi)}$$ $$BL=(BM+MT)\frac{\sin\xi}{\sin(\alpha+\xi)}=\frac{\sqrt{ax+ab}+\sqrt{ax}}{\frac{\sin\alpha}{\tan\xi}+\cos\alpha}=\frac{\sqrt{ax+ab}+\sqrt{ax}}{\sqrt{\frac{x+b}{a+x+b}}\cdot\frac{\sqrt{ax}}{x}+\sqrt{\frac{a}{a+x+b}}}=\sqrt{x(a+b+x)}=\sqrt{\frac{1}{2}\left(\sqrt{a^2+6ab+b^2}-a-b\right)\cdot\frac{1}{2}\left(\sqrt{a^2+6ab+b^2}+a+b\right)}=\sqrt{\frac{1}{4}\left(a^2+6ab+b^2-(a+b)^2\right)}=\sqrt{ab}$$
On
$\underline{\mathrm{Step\space 1}}$
We start the proof by doing some angle chasing. Please pay your attention to $\mathrm{Fig.\space 1}$, in which we have reproduced OP's sketch after introducing two auxiliary lines, namely $AT$ and $AL$. We also added the circumcircle of the quadrilateral $ABLT$ to visually confirm that it is cyclic.
Let us denote $\measuredangle ABM$ and $\measuredangle LAB$ as $\omega$ and $\phi$ respectively. Since $\measuredangle ABC$ is a right angle, $\measuredangle MBC$ is equal to $90^o-\omega$. The triangle $CMB$ is a right angle triangle. Therefore, $$\measuredangle BCM = \omega. \tag{1}$$
Consider the quadrilateral $ABLT$, the two opposite angles $\measuredangle ABC$ and $\measuredangle LTA$ of which are supplementary. This means $ABLT$ is a cyclic quadrilateral. The two angles $\measuredangle LAB$ and $\measuredangle LTB$ are subtended at the circumference of the circumcircle of $ABLT$ by the same chord $BL$. Therefore, $$\measuredangle LTB = \phi. \tag{2}$$
Since $ANT$ is a semicircle, $\measuredangle NTA$ is equal to $90^o$. As a consequence, we have $\measuredangle MTA = 90^o-\phi $. Finally, because $AMT$ is a right angle triangle, $$\measuredangle TAM = \phi. \tag{3}$$
We now have all the angles we need to proceed.
$\underline{\mathrm{Step\space 2}}$
Let the segment $MN$ be equal to $\mathrm{\mathbf{\mathit{x}}}$. Consider the right angle triangle $ABM$. We have denoted the angle at its vertex $B$, i.e, $\measuredangle ABM$, as $\omega$ and, therefore, we shall write, $$AB=\frac{AM}{\sin\left(ABM\right)}=\frac{a}{\sin\left(\omega\right)}. \tag{4}$$
Next, consider the right angle triangle $ABL$. Here, we have denoted the angle at its vertex $A$, i.e, $\measuredangle LAB$, as $\phi$. Theerefore, we can express $BL$ as $BL=AB\tan\left(LAB\right)=AB\tan\left(\phi\right)$. Once we replaced $AB$ with its value obtained above, we get, $$BL=AB\tan\left(\phi\right)=a\frac{\tan\left(\phi\right)}{\sin\left(\omega\right)}. \tag{5}$$
$\underline{\mathrm{Step\space 3}}$
Here, we try to find an expression for $\left(\frac{\tan\left(\psi\right)}{\sin \left(\omega\right)}\right)$ in terms of $a$ and $b$. Consider the right angle triangle $ABC$. As shown below, we use equation (1) to derive an expression for $AB$. $$AB=AC \sin\left(BCA\right)= AC \sin\left(BCM\right)=\left(a+b+x\right) \sin\left(\omega\right)$$
When we eliminate $AB$ from this equation using equation (4), we get, $$\frac{a}{a+b+x}=\sin^2\left(\omega\right). \tag{6}$$ The relationship given below follows from the harmonic segment containing four collinear points $A$, $M$, $N$, and $C$. $$\frac{MN}{NC}=\frac{AM}{AC}\quad \rightarrow\quad \frac{x}{b}=\frac{a}{a+b+x}.$$
Equation (6) can be used to replace the right hand side of this equation to describe $b$ in terms of $x$. $$b= \frac{x}{\sin^2\left(\omega\right)} \tag{7}$$
Consider the right angle triangle $TMN$. Its side $TM$ can be expressed in terms of $x$ using the value of $\measuredangle NTM$ given in equation (2) as, $$TM=\frac{MN}{\tan\left(NTM\right)}=\frac{x}{\tan\left(\phi\right)}. \tag{8}$$
Now, consider the right angle triangle $AMT$. This time, we find an expression for $TM$ in terms of $a$ using the value of $\measuredangle TAM$ given in equation (3). $$TM=AM\tan\left(TAM\right)=a\tan\left(\phi\right). \tag{9}$$
We use equations (8) and (9) to express $a$ as a function of $x$. $$a=\frac{x}{\tan^2\left(\phi\right)} \tag{10}$$
Now divide equation (7) by equation (10), to obtain the expression we set out to determine. $$\frac{b}{a}=\frac{\tan^2\left(\phi\right)} {\sin^2\left(\omega\right)} \quad \rightarrow \quad \frac{\tan\left(\phi\right)} {\sin\left(\omega\right)}=\sqrt{\frac{b}{a}} \tag{11}$$
$\underline{\mathrm{Conclusion}}$
Now, it remains only to substitute the value from equation (11) in equation (5). $$BL= a\frac{\tan\left(\phi\right)}{\sin\left(\omega\right)}=a\sqrt{\frac{b}{a}}=\sqrt{ab} $$
The geometric solution starts from the harmonic division and then the identification of the inscribable quadrilateral. Finally the metric relations in the right triangle and the similarity of triangles: https://twitter.com/arnoldferriv/status/1318262030068031488