jmc

Consider a triangle $ABC$ where $a=b.$



As $\angle ACB$ approaches $180^{\circ },$ $c$ approaches $2a,$ so $\frac{a^2+b^2}{c^2}$ approaches $\frac{a^2+a^2}{(2a{)}^2}=\frac{1}{2}.$ This means $M\le \frac{1}{2}.$

On the other hand, by the triangle inequality, $c<a+b,$ so $$c^2<(a+b{)}^2=a^2+2ab+b^2.$$By AM-GM, $2ab<a^2+b^2,$ so $$c^2<2a^2+2b^2.$$Hence, $$\frac{a^2+b^2}{c^2}>\frac{1}{2}.$$Therefore, the largest such constant $M$ is $\boxed{\frac{1}{2}}.$