How can I prove this problem on geometry? I need to prove the following:

If $P$ is an inner point of $\triangle ABC$, then there is a single transverse $EF$ of $\overleftrightarrow{AB}$ and $\overleftrightarrow{AC}$, where $E$ is on $\overleftrightarrow{AC}$ and $F$ is on $\overleftrightarrow{AB}$, such that P is the middle of the segment $XY$.

I tried for a longer time but nothing!! Some idea? Thanks in advance.

This answer is based on @noneEggs's suggestion.

1: Extend BP to Q such that BP = (0.5) BQ
2: (Can be skipped.) A circle (centered at P and radius = PB) is drawn cutting BP extended at Q.
3: A line through Q is drawn parallel to AB cutting AC (extended if necessary) at R.
4: Join BR.
5: A line through Q is drawn parallel to RB cutting AB at S.
6: Join SR.
According to the construction, SBRQ is a parallelogram with diagonals BQ and SR cutting each other at T.
Since the diagonals of a parallelogram bisect each other, BT = (0.5) BQ. This, together with the result in (1), means that T and P are the same point. Thus, RPS is a straight line with SP = PR.
R and S are then the required E and F respectively.
