Question about locus in rhombus. Suppose we have rhombus $ABCD$ and we have point $X$ inside our rhombus.
Now we should find locus of such points : $AXD$ + $BXC$ = $\pi$.
My attempt : obviously point of diagonals intersection is a good one.
Now suppose there is another point. So let : $d_{1} = AX$ , $d_{2} = DX$ , $d_{3} = BX$ and $d_{4} = CX$.
I figured out that : 
$cos(\alpha)=\frac{d_{1}^2 + d_{2}^2 - d_{3}^2-d_{4}^2}{d_{1}d_{2}+d_{3}d_{4}}$ but that doesn't help me at all
 A: Guess-1:
Any point on diagonals satisfy.
Proof of guess: By chasing the angles.
Guess-2:
If for some $X$ we have $AXD + BXC = \pi$ then $X$ should be on a diagonal.
Proof of guess: By "Copy paste method". "Copy the triangle $AXD$ and paste it on the side $CB$." What this means is construct the point $X'$ such that $X$ and $X'$ are on the opposite sides of $BC$ and triangles $AXD$ and $BX'C$ are identical. Now it is not hard to see $XX'$ is parallel to $CD$ and $AB$. Since we have a rhombus we get $$|XX'|=|AB|=|CD|=|BC|=|AD|$$
This was a crucial result. Second important result comes from the fact that $\angle CXB + \angle CX'B =\pi$. This follows from the fact that $AXD$ and $BX'C$ are identical and therefore $\angle CX'B = \angle DXA$. Now we can see that 
$$
X,C,X,'B \ \ \mbox{are cyclic}
$$
Therefore if $XX'$ and $BC$ intersect at $P$, we should have 
$$
|CP|\cdot|PB| = |XP| \cdot |PX'|
$$
Since $|CB|=|XX'|$, we should have either $|CP|=|XP|$ or $|CP|=|PX'|$. In each case $X$ should be on a diagonal.
