4. In $\triangle A B C$, $A B=A C=2, B C$ side has 100 different points $P_{1}, P_{2}, \cdots, P_{100}$, let $m_{i}=A P_{i}^{2}+B P_{i} \cdot P_{i} C \quad(i=1,2, \cdots$, 100) then $m_{1}+m_{2}+\cdots+m_{100}=$ $\qquad$
4. 400

Solution: Let $A D \perp B C$, as shown in the figure, then $B D=D C$. Let $B D=D C=y, D P_{i}=x$,
then
$$
\begin{aligned}
m_{i} & =A P_{i}^{2}+B P_{i} \cdot P_{i} C \\
& =A P_{i}^{2}+(y-x)(x+y) \\
& =A P_{i}^{2}-x^{2}+y^{2} \\
& =A D^{2}+y^{2} \\
& =A C^{2}=4 . \\
\therefore \quad & m_{1}+m_{2}+\cdots+m_{100}=400 .
\end{aligned}
$$