Let $AC=m$. Then in $\triangle{ABC}$, by the law of cosine, we have $$\cos{B}=\frac{a^2+b^2-m^2}{2ab}$$ And in $\triangle{ACD}$, we have $$\cos{D}=\frac{c^2+d^2-m^2}{2cd}$$ Because $ABCD$ is an inscribed quadrilateral, we have $B+D=\pi \implies \cos{B} = -\cos{D}$. It follows that $$\frac{a^2+b^2-m^2}{2ab} =\frac{m^2-c^2-d^2}{2cd} =\cos{B} = \frac{a^2 + b^2 -c^2 - d^2}{2(ab+cd)} $$ Therefore the claim hold. Note: the last step utilizes the fact that $$\frac{a}{b}=\frac{c}{d}\implies\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}$$