Why ElGamal signature is insecure with nonrandom k? In ElGamal signature scheme like:

Keygen:
$\beta:=\alpha^a\bmod p$ where $\alpha$ and $p$ are large prime numbers. Public key is $(p,\alpha,\beta)$ and secret key is $a$.

Sign:
Choosing random $k$ such that $\gcd(k,p-1)=1$, calculating $r:=a^k\bmod p$ and $s:=k^{-1}\cdot(m-a\,r)\bmod(p-1)$; then we send $(m,r,s)$

Verify:
Calculating $v_1:=\beta^r\cdot r^s\bmod p$ and $v_2:=\alpha^m\bmod p$; then if $v_1$ and $v_2$ are equal the signature verifies.

In the booklet I'm reading for cryptography course it is said that if $k$ is non-random this signature scheme is insecure. What's the reason?
If we use constant $k$ in this scheme we can find the secret key like this:

We know $(m_1,r_1,s_1)$ and $(m_2,r_2,s_2)$ for two different plaintext, as k is constant $r_1=r_2=r$ so we have this equivalence:
$$s_1k-m_1\equiv-ar\equiv s_1k-m_1\ mod\ p-1 \Rightarrow (s_1-s_2)k \equiv m_1-m_2\ mod\ p-1$$

This equivalence has $d$ answer where $d=gcd(s_1-s_2,p-1)$, $d$ is usually small so we can find $k$ using brute force on $r\equiv \alpha^k\ mod\ p$.

Now we know $k$ we can caluculate $a$ using $a.r\equiv m_1-k.s_1$ equivalence. Number of answer for this equivalence is $gcd(r,p-1)$ by comparing $\beta$ and $\alpha^a$ we can find $a$.

Now we know secret key and sign every plain text we want and that's insecure.