Squares of 4 segments In triangle $ABC$, BC=5, AC=6, and AB=7. Let $P_1, P_2, P_3, P_4$ be points on BC, in that order, so that $BP_1=P_1P_2=P_2P_3=P_3P_4=P_4C=1$. Find $AP_1^2+AP_2^2+AP_3^2+AP_4^2$. I applied Appolonius Theorem 4 times and manipulated but I am getting answer as 149. Which is incorrect. Is it a calculation error? I don't know. Please help. Thanks.
 A: By the cosine theorem, we have:
$$ BP_1^2+AP_1^2-AB^2 = 2BP_1 AP_1 \cos\theta $$
as well as:
$$ AP_1^2+CP_1^2-AC^2 = 2CP_1 AP_1 \cos(\pi-\theta) $$
where $\theta=\widehat{AP_1 B}$. Since $\cos(\pi-\theta)+\cos\theta=0$, that implies:
$$ CP_1 BP_1(CP_1+ BP_1) + (CP_1+BP_1) AP_1^2 = CP_1 AB^2 + BP_1 AC^2 $$
or:
$$ BC(CP_1 BP_1 +AP_1^2) = CP_1 AB^2 + BP_1 AC^2\tag{1}$$
that is just the statement of Stewart's theorem. 
By summing all the identities like $(1)$ that we get for $P_1,P_2,P_3,P_4$ we have:
$$BC\sum_{j=1}^{4}BP_j CP_j + BC\sum_{j=1}^{4}AP_j^2 = AB^2\sum_{j=1}^{4}CP_j + AC^2\sum_{j=1}^{4}BP_j \tag{2} $$
or:
$$ 5\left(4+6+6+4\right)+5\sum_{j=1}^{4}AP_j^2 = 10(AB^2+AC^2) = 850\tag{3}$$
from which:
$$ \sum_{j=1}^{4}AP_j^2 = 170-20 = \color{red}{150}\tag{4}$$
follows.

Also notice that the sum $\sum_{j=1}^{4}AP_j^2$ is related with the inertia moment of the segment $BC$ rotating around $A$, hence by the Huygens-Steiner's theorem such a sum depends only on $P_1 P_4^2$ and $AM^2$, where $M$ is the midpoint of $BC$.
