## Prove P,Q,A,O concyclic

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### Prove P,Q,A,O concyclic

$D$ is a arbitrary point on side $BC$ of $\triangle ABC$. The perpendicular bisector of $BD$ meets $AB$ at $P$ and the perpendicular bisector of $CD$ meets $AC$ at $Q$. If $O$ is the circumcenter of $ABC$, then prove that $P,Q,A,O$ concyclic.

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### Re: Prove P,Q,A,O concyclic

Let $E,F$ be the midpoints of $BD,DC$ respectively .
$AD' \perp BC$ ($D' \in BC$). $E',F',P',Q'$ are midpoints of $BD',D'C,AB,AC$ respectively .

Note that , $EE'=FF'$

Now $EE'=PP'\cos {B}$ and $FF'=QQ' \cos {C}$
so $\displaystyle \frac {PP'}{QQ'}=\frac{\cos {C}}{\cos {B}}=\frac {OP'}{OQ'}$ and $\angle {OP"P}=\angle {OQ'Q}$
imply $\triangle OP'P ~ \triangle OQ'Q$ so $\angle {OPP'}=\angle {OQQ'}$
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### Re: Prove P,Q,A,O concyclic

Let us denote $\angle BAC,\angle CBA, \angle ACB$ by $\angle A,\angle B,\angle C$, respectively. Also denote $R$, the circumradii of $ABC$.
Step 1:
$\dfrac {\sec \angle B}{\tan \angle B+\tan \angle C}=\dfrac {\cos \angle C}{\sin \angle A}$

Proof: Nothing but a muscle exercise.

Denote $PS\cap QD=X$, $QR\cap PD=Y$. Notice that $\angle OBP=\angle QXP=|\pi/2-\angle C|$

$BP=\frac {BD}{2}\cdot \sec \angle B$
$XP=\frac {BD}{2}\cdot (\tan \angle B+ \tan \angle C)$
$XQ=\frac {BC}{2}\cdot \sec \angle C$

Therefore,
\begin{align*}\frac {BP}{XP}&=\frac {\sec \angle B}{\tan \angle B+\tan \angle C}\\ &=\frac {\cos \angle C}{\sin \angle A}\\ &=\frac {1}{\sin \angle A\cdot \sec \angle C}\\ &=\frac {2R}{BC\cdot \sec \angle C}\\ &=\frac {R}{XQ}\\ &=\frac {BO}{XQ}\end{align*}

Hence $XPQ\sim BPO \Longrightarrow \angle OPQ=\angle BPX=\angle OAQ\Longrightarrow A,P,O,Q$ concyclic.
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