First, let's show that if $MN\perp AB$, then $AM^2-BM^2=AN^2-BN^2$. For this, let $E$ be the intersection point of $AB$ and $MN$. Then by the Pythagorean theorem, we have $$ \begin{array}{} AM^2 = AE^2 + ME^2, BM^2 = BE^2 + ME^2 \implies AM^2 - AN^2 = AE^2 - BE^2\\ AN^2 = AE^2 + NE^2, BN^2 = BE^2 + NE^2 \implies AN^2 - AN^2 = AE^2 - BE^2 \end{array} $$ $$\therefore\quad AM^2-BM^2=AN^2-BN^2$$ Next, if $AM^2-BM^2=AN^2-BN^2$, let's show that $AM\perp MN$ using law of cosines. For convenience, let $\angle{AME}=\alpha$ and $\angle{BME}=\pi-\alpha$. $$ \begin{align} AM^2 &= AE^2 + ME^2 - 2 AE\cdot ME\cdot cos\alpha\\ BM^2 &= BE^2 + ME^2 - 2 BE\cdot ME\cdot \cos(\pi-\alpha) = BE^2 + ME^2 + 2 BE\cdot ME\cdot \cos\alpha \end{align} $$ $$\therefore\quad AM^2 - BM^2 = AE^2 - BE^2 -2 (AE+BE)ME\cos\alpha=AE^2-BE^2-2AB\cdot ME\cos\alpha$$ Similarly, we must also have $$AN^2 - BN^2 =AE^2-BE^2-2AB\cdot NE\cos\alpha$$ Setting these two equal will result in $$ME\cos\alpha = NE\cos\alpha$$ This holds only if $\cos\alpha = 0$ which implies $\alpha=90^\circ$ or $AB\perp MN$.