Prove relationship among sides of 30-60-90 triangle only with median to hypotenuse How to prove $BC:AC:BA = 1:\sqrt{3}:2$ using this diagram with $CM$ being the median to the hypothenuse $BA$ and without adding any more lines?

I tried to show that $\Delta BCM$ is equilateral which would yield $BC:BA = 1:2$ and by Pythagoras theorem $BC:AC:BA = 1:\sqrt{3}:2$
To show $\Delta BCM$ is equilateral I tried showing angles $x = m = 60$ so I came up with these equations utilizing the constant sum of angles in a triangle and exterior angles:

But I think this doesn't get me far.
BTW: This problem is being asked in AoPS - Volume 1: The Basics in order to derive values for $\sin$ and $\cos$ for 30° and 60°:

 A: I may have found an answer myself: Knowing that the median to the hypotenuse of a right triangle is half the hypotenuse, we can immediately conclude $\Delta BCM$ is equilateral and thus deriving the ratio of the sides like @Prem did.
Showing this fact involves drawing the circumcircle of $\Delta BCA$:

Since $\angle C$ must be right, it needs to cut off a $180°$ arc implying $BA$ is a diameter of the circumcircle and M being its midpoint. Hence $MC, MB, MA$ are all radii of the circumcircle and thus equal.
A: An alternative is to use coordinates. Assume $x$ axis is along $CA$ and $y$ axis is along $CB$. Then the coordinates of the corners are $C=(0,0)$, $A=(b,0)$, $B=(0,a)$. The middle of $AB$ is then at $$M=\frac12((b,0)+(0,a))=\left(\frac b2,\frac a2\right)$$
Then $$CM^2=\frac {b^2}4+\frac{a^2}4=MB^2=\frac {c^2}4$$
This means that $\triangle CMB$ is isosceles. Since $\angle CBM=\angle CBA=60^\circ$, the triangle is equilateral. Therefore $$a=\frac c2$$ Using Pythagoras, $$a^2+b^2=c^2\\b=c^2-a^2=(2a)^2-a^2=3a^2$$
So $a:b:c=1:\sqrt 3:2$.
A: What this Picture is trying to tell :
---- We are given right-angle triangle $ABC$ with $90^0-60^0-30^0$.
---- Draw a line from the right-angle at $60^0$ to meet the hypotenuse $AB$ at $M$.
Nothing more , nothing less !!
This is what we get :

In triangle $CBM$ : $CBM = 60^0$ , $BCM = 60^0$ , hence $CMB = 60^0$ , all sides are $a$.
Also $CM=AM$ , hence hypotenuse is $2a$.
This is what we know now :

Pythagoras tells us that $a^2+b^2 = (2a)^2$.
Now we eliminate $b$ & get the ratios.
$b^2 = 4a^2-a^2 = 3a^2$
$b = \sqrt{3}a$
The sides are $a,2a,\sqrt{3}a$.
The sides are in the ratio $1,2,\sqrt{3}$.
No new lines Drawn !
