If $a, b, c, d \in \Bbb R$ and $ab+cd, ad+bc \neq 0$, prove $\vert(a^2+b^2-c^2-d^2)/(ab+cd)\vert \lt 2 ⇒\vert(a^2-b^2-c^2+d^2)/(ad+bc)\vert \lt 2$

Question

If $\vert (a^2+b^2-c^2-d^2)/(ab+cd)\vert \lt 2$ then there exists an angle $\theta$ between $0$ and $\pi$ such that $2\cos\theta=(a^2+b^2-c^2-d^2)/(ab+cd)$. Then $a^2+b^2-2ab\cos\theta=c^2+d^2+2cd\cos\theta$. I used law of cosines. There are three cases $abcd=0$, $abcd \gt 0$ and $abcd \lt 0$. I don't write the details here. I want to know if there is a nongeometric solution to this problem.