Squares on the sides of a regular pentagon On each side of a regular pentagon a square lying outside the pentagon is constructed. (see the picture below)
$X_1,..., X_5$ are the centers of the squares.
$P$ and $Q$ are points of intersection of $X_5X_2$ with $X_1X_4$ and $X_1X_3$ respectively.
$O$ is the center of the pentagon.
Prove that the circumcircle of $OPQ$ touches $X_1X_4$ and $X_1X_3$


Note that, by the symmetry of the picture, $\angle OQP= \angle OPQ= \angle OPX_4$. Consider $\triangle OQP$ and the circumcircle. We have that the inscribed angle over $OP$, $\angle OQP$, is equal to the angle between $OP$ and $PX_4$, so $PX_4$ is the tangent to the circumcircle.
