Solve system of 2 equations with 3 unknowns

We are given a triangle $ABC$ with sides $a, b, c$ respectively and for which the following relationships hold: $a^2+bc\sqrt 3 = b^2+c^2$,

$c^2+ba = a^2+b^2$

We want to prove that angle $B$ is right.

I am trying to express sides $b$ and $c$ in relation to $a$ and then prove that they satisfy the Pythagorean theorem.

By combining the two equations, I am getting:

$b = \frac {c\sqrt 3+1}{2}$

Then I am plugging this expression into the second equation, in order to eliminate $b$:

$4a^2-2a(c\sqrt 3+1)-c^2+2c \sqrt 3 +1=0$

and I must now solve for $a$ in relation to $c$ but I am getting a complex expression, which, by no means, satisfies Pythagorean.

I input the two initial equations in Wolfram and it gives as a solution (apart from the ones which are rejected): $b=2a$ and $c=a\sqrt 3$ which clearly satisfy Pythagorean, because $b^2 = 4a^2 = c^2+a^2$.

Any ideas? Thank you!!

Comparing the first equation with cosine rule:

$$a^2=b^2+c^2-2bc\cos A,$$ $$a^2=b^2+c^2-\sqrt3bc,$$ $$\implies\cos A=\frac{\sqrt3}2\implies \measuredangle A=30^\circ.$$

Now using the two given equations we get, $2b=\sqrt3c+a$.

Applying sin rule, $$2\sin B=\sqrt3\sin C+\sin A$$ $$2\sin B=\sqrt3\sin(150^\circ-B)+\sin30^\circ$$

Continue this to solve for $\measuredangle B$ and you will get the answer.

Alternate solution: As I've mentioned previously, it should be $2b=\sqrt3c+a$, not $2b=\sqrt3c+\color{red}1$.

So if you substitute the value of $b$ in the second relation in terms of $a$ and $c$, you will get $c^2=3a^2$. Hence, the problem is solved.