<s>[INST] What's the measure of the segment $BC$ in the rhombus below? For reference: Given a rhombus $ABCD$, on $BC$
mark the point $P$ such that : $BP= 3PC$ and $AP^2+ 3DP^2 = 38$.
Calculate $BC$.(answer: $2\sqrt2$)
My progress:

$BP = 3CP\\
AP^2+3DP^2 = 38\\
AB=BC=CD=AD$
Th. Stewart:
$\triangle ABC:\\
AC^2.BP+AB^2.CP=AP^2BC+BC.CP.BP\\
AC^2. 3CP+AB^2,CP = BC(AP^2+3CP^2)\\
\boxed{CP(3AC^2+AB^2) = BC(AP^2+3CP^2)}(I)\\
\triangle DBC:\\
CD^2.BP+BD^2CP=DP^2.BC+BC.BP.CP\\
CD^2.3CP+BD^2.CP=BC(DP^2+3CP^2)\\
\boxed{CP(3CD^2+BD^2) = BC(DP^2+3CP^2}(II)$
(I)+(II):
$\boxed{CP(3(AC^2+CD^2)+AB^2+BD^2) = BC(AP^2+DP^2+6CP^2)(III)}$
...??
 [/INST] I would have simply placed the figure on a coordinate plane such that $$\begin{align}
C &= (x,0), \\ 
B &= (0,y), \\
A &= (-x,0), \\
D &= (0,-y), \\
\end{align}
$$
hence $$P = (\tfrac{3}{4}x, \tfrac{1}{4}y),$$
and
$$AP^2 = \left(\tfrac{7}{4} x\right)^2 + \left(\tfrac{1}{4}y\right)^2 = \frac{49x^2 + y^2}{16}, \\
DP^2 = \left(\tfrac{3}{4}x\right)^2 + \left(\tfrac{5}{4}y\right)^2 = \frac{9x^2 + 25y^2}{16}.$$
Thus $$38 = AP^2 + 3DP^2 = \frac{76(x^2 + y^2)}{16} = \frac{19}{4}(x^2 + y^2) = \frac{19}{4}BC^2,$$ from which the result immediately follows.
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