BdMO 2017 National Round Secondary 9

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Kazi_Zareer
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BdMO 2017 National Round Secondary 9

Unread post by Kazi_Zareer » Fri Feb 10, 2017 9:24 pm

In a cyclic quadrilateral $ABCD$ with circumcenter $O,$ the lines $BC$ and $AD$ intersect at $E.$ The lines $AB$ and $CD$ intersect at $F.$ A point $P$ satisfying $\angle EPD = \angle FPD = \angle BAD$ is chosen inside of $ABCD.$ The line $FO$ intersects the lines $AD,EP,BC$ at $X,Q,Y$ respectively. Also $\angle DQX = \angle CQY.$ What is the $\angle AEB$?
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joydip
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Re: BdMO 2017 National Round Secondary 9

Unread post by joydip » Tue Feb 14, 2017 9:08 pm

Let $QO$ meet $(DQC)$ again at $O'$, then $\angle DQX = \angle CQY \Rightarrow DO'=O'C$ implying $O=O'$. Then $FD .FC=FQ.FO \Rightarrow Q$ is the invertion of $F$ wrt $(ABCD)$ ,by brocard's theorm $EQ$ is the polar of $F$ wrt $(ABCD) \Rightarrow -1=(EF,EP;ED,EC)=(PF,PE;PD,PC)$. $\angle EPD = \angle FPD$ gives $\angle DPC=90^\circ$. $\angle DPE=\angle BAD=\angle DCE \Rightarrow DPCE$ is cyclic $ \Rightarrow \angle AEB=180^\circ-\angle DPC=90^\circ$
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NOBODY
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Re: BdMO 2017 National Round Secondary 9

Unread post by NOBODY » Thu Feb 16, 2017 9:49 pm

How to draw a good diagram for this? I couldnt construct the point P with hands..tried for a long time..

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