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Let ABCD be a cyclic quadrilateral. Let r and s be the lines obtained by reflecting AB through the angle bisectors of $\angle CAD$ and $\angle CBD$, respectively. Let P be the intersection of r and s and let O be the center of ABCD. Problem: Show that OP is perpendicular to DC. DRAWING. I have showed that the bisectors cut at the midpoint of arc CD, and since a perpendicular line to CD that passes through O must cut CD in its midpoint, so I want to show that CP=PD.

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Let $E$ be the intersection of circle $ABCD$ and the reflection of $AB$ through $s$. It follows that $\angle ABD=\angle CBE$. So the two arcs $AD$ and $CE$ are equal and we have $AE\parallel CD$.

Similarly, let $F$ be the intersection of circle $ABCD$ and the reflection of $AB$ through $r$, then $BF\parallel CD$.

It follows that $AF$ and $BE$ intersect on the common perpendicular bisector of $CD$, $AE$, and $BF$.

Hint: Show that $P$ is the midpoint of the arc $CD$ which does not contain $A$ (and $B$). In other words, if $Q$ denotes that point, then $AQ$ is the angle bisector of $\angle CAD$ and $BQ$ is the angle bisector of $\angle CBD$.

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  • $\begingroup$ I think you are confusing P with the intersection of the angle bisectors, which trivially by angle chasing cut at the midpoint of arc CD. $\endgroup$ – problembuster Jul 28 '16 at 7:41
  • $\begingroup$ P is the intersection of the reflections of AB through the angle bisectors of ∠CAD and ∠CBD. $\endgroup$ – problembuster Jul 28 '16 at 7:43
  • $\begingroup$ Ah, right. What was I thinking :-( $\endgroup$ – Quang Hoang Jul 28 '16 at 8:00
  • $\begingroup$ @problembuster How about this fix? $\endgroup$ – Quang Hoang Jul 28 '16 at 8:31

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