As quadrilateral $BMGP$ is cyclic, $AP.AB=AG.AM$ Hence, $\frac{AB^2}{2} = \frac{2AM^2}{3} \implies 2AM = \sqrt{3} AB \implies AM=BN$ Hence, $\angle A = \angle B \implies AC=BC$ Let $AN=CN=CM=BM=x$ In $\triangle ABC$, Apollonius theorem $\implies 2AM^2 + 2BM^2 = AB^2 + AC^2 $ $\frac{AB^2}{2} = \frac{2AM^2}{3} \implies AB^2 = \frac{4AM^2}{3} \implies \frac{2AM^2}{3} = 2x^2 \implies AM^2 = 3x^2$ In $\triangle ABC$, Apollonius theorem $\implies AC^2 + BC^2 = 2CP^2 + 2AP^2 $ $AC^2 = BC^2 = 4x^2 , AP^2 = \frac{AB^2}{4} = \frac{AM^2}{3} \implies CP^2 = 4x^2 - \frac{AM^2}{3}$ $AM^2 = 3x^2 \implies CP^2 = 3x^2$ Hence, $CP^2 = AM^2 = BN^2 \implies \triangle ABC$ is equilateral