Prove that the segment $BE$ bisects $AC$ Points $A,B,C,D,E$ lie on a circle $ω$ and point $P$ lies outside the circle. The given points are such that (i) lines $(P B)$ and $(P D)$ are tangent to $ω$, (ii) $P, A, C$ are collinear, and (iii) $DE ∥ AC$.
Prove that $[BE]$ bisects $[AC]$.
The issue that I have with this problem is that I don't what should I prove to get that $BE$ bisects $AC$. Maybe proving that the intersection between $(BE)$ and $(AC)$ (let's call it $I$) is inside $\omega$. Which is equivalent to proving $$OI<OA \quad \text{where O is the center of }  \omega.$$
Would that work?


Since $PB$ and $PD$ are tangents to the circle $\omega$ at the points $B$ and $D$ respectively, the two triangles $BOP$ and $DOP$ are congruent, right-angled triangles where $$\angle \, OBP = \angle\, ODP = 90$$
Moreover, by congruence, $$\angle \, BOP = \angle\, DOP = \alpha$$
which means that $$\angle \,  BOD = 2\, \alpha$$
But then
$$\angle\, BED = \frac{1}{2} \, \angle \, BOD = \alpha$$
because $\angle \,  BOD$ is a central angle and $\angle \,  BED$ is inscribed. However, $AC \, || \, DE$ which by the collinearity of $A, C, P$ means that
$$PC \, || \, DE$$ and therefore
$$\angle \, BIP = \angle \, BED = \alpha$$
Thus, we conclude that $$\angle \, BIP = \alpha = \angle \, BOP$$ which means that the quadrilateral $BOIP$ is cyclic (in fact the points $B, I, O, D, P$ lie on the same circle.) Consequently,
$$\angle \, OIP = \angle\, OBP = 90$$ which means that $$OI \perp AP$$ however, $C$ lies on AP, so $$OI \perp AC$$ and since $AC$ is a chord in the circle $\omega$ with $O$ its center, the point $I$ must be the midpoint of the segment $AC$, i.e. $BE$ bisects $AC$.
