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):
$\boxed{CP(3(AC^2+CD^2)+AB^2+BD^2) = BC(AP^2+DP^2+6CP^2)(III)}$
...??
 A: $ \small AB^2 \cdot CP + AC^2 \cdot 3 CP = 4CP \cdot(AP^2 + 3 CP^2)$
As $ \displaystyle \small AB = BC \text { and } BC = 4 CP$,
$ \displaystyle \small BC^2 + 3 AC^2 = 4 (AP^2 + \frac {3 BC^2}{16})$
$ \displaystyle \small BC^2 + 12 AC^2 = 16AP^2 \tag1$
Similarly,
$ \displaystyle \small BD^2 \cdot CP + CD^2 \cdot 3CP = 4 CP \cdot (DP^2 + 3CP^2)$
$ \displaystyle \small BD^2 + 3 BC^2 = 4 (DP^2 + \frac{3 BC^2}{16})$
$ \displaystyle \small 4BD^2 + 9BC^2 = 16 DP^2 \tag2$
By $(1) + 3 \cdot (2), ~$
$ \displaystyle \small 28BC^2 + 12 (AC^2 + BD^2) = 16 \cdot 38 $
Note that $ \displaystyle \small AC^2 + BD^2 = 4 BC^2$ given a rhombus.
So, $ \displaystyle \small 76 BC^2 = 16 \cdot 38 \implies BC^2 = 8$
$ \therefore BC = 2\sqrt2$
A: 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.
