Forgive my popping this thread, I really want an answer for this problem

An answer? It is not true! By interchanging, we must have $PQ\bot AD$ as well! Best regards, sunken rock

sunken rock wrote: An answer? It is not true! By interchanging, we must have $PQ\bot AD$ as well! That's the point !!! It' s obvious that if PQ is perpendicular to BC, since EF bisects $\angle BEC$, it sends $\triangle FAB$ to $\triangle FCD$ and they're congruent! I have no idea how to use the second condition, the "$DQ \cdot BP=AQ \cdot CP$" must be very strict if this statement is true, I tried basic Menelause and Harmonic Division to get an equation form like $\frac{PX}{PY}=\frac{AQ}{DQ}$ where X,Y are 2 points on BP,CP, respectively but I failed to get anything...... The original version in Chinese posted in a Chinese Maths Forum is the same as this one, maybe I just haven't got the key to this problem?

the triangles FAB and FCD are similar. so the angle AFB is congruent with DFC so the triangles AFC and DFB are congruent so AC=BD and the angle BDC is congruent with the angle BAC so ABCD is cyclic with AC=BD so ABCD has AD is parallel with BC. if AB is not parallel with CD because in the case AB parallel to DC F can't be defined. so F is the circumcircle of ABCD. and EA=EC so E,F are on the perpendicular bisector of BC so EF is perpendicular on BC. where is the mistake?????????????? I didn't use $ DQ\cdot BP=AQ\cdot CP $. HELP!!!!!!!

I'm not sure, do you mean AC=BD implies "$\angle BDC = \angle BAC$" or "$\triangle BDC$ is congruent to $\triangle BAC$" ? But neither of them is obvious, how do you get it?

$ \angle BDC =\angle BAC $ this is because $ \angle FAC + \angle FAB = \angle BDC $,$ \angle BDF + \angle FDC $ and $ \angle FAC=\angle BDF $,$ \angle FAC= \angle FDC $.

I am very confused

anonymouslonely wrote: $ \angle BDC =\angle BAC $ this is because $ \angle FAC + \angle FAB = \angle BDC $,$ \angle BDF + \angle FDC $ and $ \angle FAC=\angle BDF $,$ \angle FAC= \angle FDC $. for spiral similarity, you should have $ \angle FAC=\angle FBD $ not $\angle FDB$

thank you very much,genxium. I have tried an inversion on F and I think I can obtain something nice.

I will finish my idea and if I will obtain something I will post.

Firstly，$AC=BD$，$PQ$ is the interior bisector of $\angle BEC$ . $AQ \cap BP=S,DQ \cap CP=T$，then $S,E,T$ are collinear，and $\frac{sin(\angle CET)}{sin(\angle DET)}=\frac{DE}{CE} \cdot \frac{BE}{AE}=m$ Using $DQ\cdot BP=AQ\cdot CP$，note $ABSE,CDTE$ ，We get $\frac{AS}{BS} \cdot \frac{CT}{DT}=\frac{sin^2(\angle CET)}{sin^2(\angle DET)}=m^2$ Let $AB \cap CD=R$，$X,Y$ on $AB,CD$ such that $RAXB,RDYC$ make harmonic division，also Using $DQ\cdot BP=AQ\cdot CP$，we get $\frac{AS}{BS} \cdot \frac{CT}{DT}=(\frac{RD}{RC} \cdot \frac{CE}{DE}) \cdot (\frac{RB}{RA} \cdot \frac{AE}{BE})=\frac{sin^2(\angle DEY)}{sin^2(\angle CEY)}=n^2$ . So $m=n$，$(\frac{DE}{CE} \cdot \frac{BE}{AE})^2=(\frac{RD}{RC} \cdot \frac{CE}{DE}) \cdot (\frac{RB}{RA} \cdot \frac{AE}{BE})$ i.e. $m^3=\frac{RD\cdot RB}{RA \cdot RC}=\frac{sin(\angle RED)}{sin(\angle REA)}=\frac{sin(\angle DEY)}{sin(\angle CEY)}=n$，so $m=n=1$，$AE \cdot CE=BE \cdot DE$，(or use Gauss's Line) So $ABCD$ is a an isosceles trapezoid and $AD$ must parallel to $BC$，so $PQ \perp BC$ .

According to a scanned copy of the test somebody gave me, $EF\cap AD=Q$ and $EF\cap BC=P$ (the condition $DQ\cdot BP=AQ\cdot CP$ is correct with the new notation). The solution for this is simple. Clearly the first condition $MF/AB=NF/CD$ combined with the fact that $FA=FB$ and $FC=FD$ implies that $F$ is the spiral center taking $AB$ to $CD$. In particular, $AFEB$ and $DFEC$ are cyclic, so $\angle{BEP}=\angle{BAF}=\angle{FDC}=\angle{PEC}$. By the angle bisector theorem, $BE/CE=BP/CP=AQ/DQ=AE/DE$, so $ABCD$ is cyclic. Thus $\angle{FAC}=\angle{FAB}-\angle{CAB}=\angle{CDF}-\angle{CDB}=\angle{BDF}=\angle{ACF}$, so $FA=FC$ and similarly $FB=FD$. But $FA=FB$ and $FC=FD$, so $F$ is the center of $(ABCD)$. Because $F$ takes $AB$ to $CD$, $AB=CD$, so $ABCD$ is an isosceles trapezoid, from which the conclusion follows easily.

math154 wrote: According to a scanned copy of the test somebody gave me, $EF\cap AD=Q$ and $EF\cap BC=P$ (the condition $DQ\cdot BP=AQ\cdot CP$ is correct with the new notation). The solution for this is simple. Clearly the first condition $MF/AB=NF/CD$ combined with the fact that $FA=FB$ and $FC=FD$ implies that $F$ is the spiral center taking $AB$ to $CD$. In particular, $AFEB$ and $DFEC$ are cyclic, so $\angle{BEP}=\angle{BAF}=\angle{FDC}=\angle{PEC}$. By the angle bisector theorem, $BE/CE=BP/CP=AQ/DQ=AE/DE$, so $ABCD$ is cyclic. Thus $\angle{FAC}=\angle{FAB}-\angle{CAB}=\angle{CDF}-\angle{CDB}=\angle{BDF}=\angle{ACF}$, so $FA=FC$ and similarly $FB=FD$. But $FA=FB$ and $FC=FD$, so $F$ is the center of $(ABCD)$. Because $F$ takes $AB$ to $CD$, $AB=CD$, so $ABCD$ is an isosceles trapezoid, from which the conclusion follows easily. Are you sure about this new notation? Because once P,Q exchanged, the problem is much easier

genxium wrote: Are you sure about this new notation? Because once P,Q exchanged, the problem is much easier I'm just going by this...

WOAH .... i was busy just then trying to solve this problem for the last two days and now i know that the statement isn't correct .... trying every aspect of the first equality like $ MN $ orthogonal to the bisector of angle $ CEB $ . Well .... taking the wrong problem statement .... if we ignore the first equality and instead assume that the quadrilateral is cyclic .... and let the $ P $ and $ Q $ be as in the wrong picture so they determine the bisector .... then one can use the second equality to show that this quadrilateral is a rectangle ....

So sorry,edited.

DQ/AQ=CP/DP implies that AB//CD and MF/AB=NF/CD implies that <FAB=<FCD. Since <EAB=<ECD, E and F must coincide. Hence ABCD is an isosceles trapezoid with AB//CD. But PQ is not perpendicular to BC. Let midponts of AB,CD be Q',P' respectively ,then P'Q' is perpendicular to AB and CD.