two angles are equal

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two angles are equal

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Source: 2011-2012 china second round,problem 1

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#1 h Oct 30, 2011, 6:09 AM • 2 Y

Y by Adventure10, Mango247

Let $P,Q$ be the midpoints of diagonals $AC,BD$ in cyclic quadrilateral $ABCD$ . If $\angle BPA=\angle DPA$ , prove that $\angle AQB=\angle CQB$ .

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#2 h Oct 30, 2011, 9:22 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

nice problem....consider $ABCD$ inscribed in circle $\omega$ with center $O$ . also $E$ is the intersection point of
$AC,BD$ and $P'$ and $Q'$ are the intersection point of $OP$ and $OQ$ with $BD$ and $AC$ ,respectively.
$OP \perp AC , OQ \perp BD, \angle APD=\angle APB$ so $(BD,EP')=-1$ and it's sufficien to prove that $(AC,EQ')=-1$ .
$(BD,EP')=-1$ so $E$ is on polar of point $P'$ . and $AC\perp OP'$ so $AC$ is polar of $P'$ . so $Q'$ is on polar of $P'$ $\Rightarrow$ $P'$ is on polar of $Q'$ . and because $BD\perp OQ'$ so $BD$ is polar of $Q'$ ..then we have what we wanted!! $(AC,EQ')=-1$

This post has been edited 1 time. Last edited by goldeneagle, Oct 30, 2011, 9:25 AM

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#3 h Oct 30, 2011, 9:23 AM • 2 Y

Y by Adventure10, Mango247

You should post in Geometry forum

It's a easy problem and it didn't took me 3 minutes in the exam

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#4 h Nov 20, 2011, 5:29 AM • 1 Y

Y by Adventure10

actually,it's just exactly the same as $2007$ Singapore TST:see here
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=224939&

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#5 h Nov 20, 2011, 3:48 PM • 2 Y

Y by Adventure10 and 1 other user

Produce $BP$ to intersect the circumcircle at $E$ ; by symmetry, the arcs $AE$ and $CD$ are equal, hence $DE\parallel AC$ , i.e. $BD$ is symmedian of $\triangle ABC$ , consequently $ABCD$ is harmonic quad, so $AB\cdot CD=BC\cdot AD=BD\cdot CP=AC\cdot BQ$ , so we easily find $\triangle ABQ\sim \triangle ACD$ , so $\angle AQD=\angle ABC$ ; we also notice $\triangle DQC\sim \triangle ABC$ , with $\angle DQC=\angle ABC$ , done.

Best regards,
sunken rock

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