Inside a square $ABCD$ point $P$ is marked, and on the sides $AB$, $BC$, $CD$ and $DA$ points $K,L,M$ and $N$ are chosen respectively. Lines $KP,LP,MP$ and $NP$ intersect sides $CD,DA,AB$ and $BC$ at points $K_1, L_1, M_1$ and $N_1$ respectively. It turned out that $$\frac{KP}{PK_1}+\frac{LP}{PL_1}+\frac{MP}{PM_1}+\frac{NP}{PN_1}=4$$Prove that $KP+LP+MP+NP=K_1P+L_1P+M_1P+N_1P$.