<Qua-2025-780>, hint

Let A(0 ,a), B(-b,0), C(0,-a), D( b,0)

and intersection of AC with BD be O(0,0).

Let AM/MB=x/1-x, DP/PC=y/1-y, BP/PC=(x+y)/(1-(x+y)

Then M(-xb, (1-x)a), P(1-y)b, -ya), N(1-(x+y)(-b), (x+y)(-a))

G(x)=1/3 {-xb+(1-y)b+(x+y)b-b}={0}. Hence G lies on AC