What's the measure of the segment $BC$ in the rhombus below?

Question

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):

$\boxed{CP(3(AC^2+CD^2)+AB^2+BD^2) = BC(AP^2+DP^2+6CP^2)(III)}$

...??

Answer 1

$ \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$

Answer 2

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.