Question 76
A point $P$ is marked inside the square $ABCD$, and the points $K$, $L$, $M$ and $N$ are marked on its sides $AB$, $BC$, $CD$ and $DA$ respectively. The lines $KP$, $LP$, $MP$ and $NP$ intersect the sides $CD$, $DA$, $AB$ and $BC$ at the 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 $KP + LP + MP + NP = K_1P + L_1P + M_1P + N_1P$.
$$
\frac{KP}{PK_1} + \frac{LP}{PL_1} + \frac{MP}{PM_1} + \frac{NP}{PN_1} = 4.
$$
Prove that $KP + LP + MP + NP = K_1P + L_1P + M_1P + N_1P$.