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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?

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PNT
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  • Since points $B$ and $E$ lie in different created by the line $PCA$ parts of the circle, $BE$ must intersect $AC$. – user Feb 10 '22 at 20:19
  • That looks like a generalization. Anyways I need a proof @user – PNT Feb 10 '22 at 20:22
  • https://artofproblemsolving.com/wiki/index.php/2011_USAJMO_Problems/Problem_5 – ACB Feb 11 '22 at 04:11
  • I would generalize this in the following way: let $\cal F$ be a convex figure and $A$ and $B$ be two points on its boundary. Then the segment $AB$ splits $\cal F$ in two parts. Claim: if the points $C,D$ of $\cal F$ are in the different parts created by $AB$ then the segments $AB$ and $CD$ intersect. – user Feb 11 '22 at 08:51

1 Answers1

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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 $AC$.

Futurologist
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