Follow-up to Find $\angle{FPC}$: impossible question?

Question

I have asked my friend about this question, and he now replied that is was a rhombus, not a parallelogram. The question is now:

Shape $ABCD$ is a rhombus.
$AE = EB$
$BF = FC$
Find $\angle{FPC}$.

Now it should be possible to solve. Here is what I did:

Since $ABCD$ is a rhombus, $AB = BC$, which means $EB = BF$, because $AE = EB$ and $BF = FC$. $EB = BF$ leads to $\angle{BEF} = \angle{EFB}$, and since we know that $\angle{EBF} = 70^{\circ}$, $\angle{BEF} = \frac{180 - 70}2$. Because $\angle{BEF} = \angle{FPC}$, the answer, $\angle{FPC}$, is $55^{\circ}$.

Wow, much easier. Now the question is: Is my proof/explanation correct?

Answer 1

You are correct, but you haven't explained why $\angle BEF=\angle FPC$.

The key to this is that because $F$ is the midpoint of $BC$, point $F$ must lie directly below the midpoint of $EP$. So $\triangle EPF$ is isosceles with $\angle FEP = \angle EPF$.