Whats is the relation between $AM$ and $BC$ in the question below?

Question

For reference: (exact copy of the question) In a right triangle ABC, straight at $"B"$, the median $AM$ is drawn. If the measurement of angle $AMB$ is twice the measurement of angle $A$, calculate: $\frac{AM}{BC}$

My progress..By trigonometry I decided, I would like the resolution by geometry, if possible..

$\triangle ABM: tan2\alpha=\frac{h}{x}(1)\\ \triangle ABC:tan\alpha=\frac{2x}{h}(2)\\ (1)x(2):tan2\alpha \cdot tan\alpha=2\\ \frac{2tan\alpha}{1−tan^2\alpha}\cdot tan⁡α=2\implies tg \alpha = \frac{\sqrt2}{2}(3)\\ (3)in(2):\\ \frac{\sqrt2}{2}=\frac{2x}{h}\implies h = 2\sqrt2x\\ (T.Pit)ABM:\\ x^2+(2\sqrt2x)^2=AM^2 \implies 3x=AM\\ \therefore \frac{AM}{BC} =\frac{3x}{2x}=\frac{3}{2}$

Answer 1

I will solve it one of the ways and show another way to go about it but leave it for you to fill the details.

First way -

Say point $D$ is reflection of $C$ about $AB$ and $N$ is reflection of $M$ about $AB$.

We know, $\angle BAM = 90^\circ - 2 \alpha$

So, $\angle DAM = \alpha + (90^\circ - 2 \alpha) = 90^\circ - \alpha$

Also, $\angle ADM = \angle ACB = 90^\circ - \alpha$

And it follows that $\triangle AMD$ is isosceles with $AM = DM = 3 x$

So, $ \displaystyle \frac{AM}{BC} = \frac{3 x}{2 x} = \frac{3}{2}$

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Second way -

Midpoint of $AC$ (point $N$) is circumcenter of the right triangle $\triangle ABC$. Intersection $G$ of $AM$ and $BN$ is the centroid, so $AG:GM = ?$. Now can you show in $\triangle BGM$, $BM = GM$ and can you finish the proof?

Answer 2

Construct bisector of $\angle AMB$. Then draw a perpendicular from $N$ to $AM$. It follows that $\triangle MON \cong \triangle MBN \implies MO=x$.
Also, $\triangle MBN \sim \triangle ABC \implies \frac{BN}{x}=\frac{2x}{h} \implies BN=\frac{2x^2}{h} $
Finally, $\triangle AON \sim \triangle ABM \implies \frac{AO}{h}=\frac{ON}{x}=\frac{2x^2}{xh} \implies AO=2x \implies AM=3x$