Given a rhombus $ABCD$. The points $E$ and $F$ are the midpoints of the line segments $AB$ and $BC$ respectively. A point $P$ satisfies the following conditions: $PA=PF$; $PE=PC$. How can I prove that the point $P$ lies on line $BD$?
Suggestions:
1) It may be possible to use Ceva's theorem here. But what triangles should I choose for it?
2) The point $P$ is the intersection point of the midpoint perpendiculars of segments $AF$ and $EC$.
Observe reflection across $BD$. It takes $F$ to $E$ and $A$ to $C$. So it takes segment bisector $k$ to $m$. So if $m$ and $k$ intersect then they intersect at axis of symmetry $BD$.
Anyway, you can also prove your problem by introduction of the coordinate system....