WindyLi wrote: In the quadrilateral $ABCD$, $AB=CD$, $CB=CD$, $\angle ABC =90^\circ$. $E$, $F$ are on $AB$, $AD$ and $P$, $Q$ are on $EF$($P$ is between $E, Q$), satisfy $\frac{AE}{EP}=\frac{AF}{FQ}$. $X, Y$ are on $CP, CQ$ that satisfy $BX \perp CP, DY \perp CQ$. Prove that $X, P, Q, Y$ are concyclic. $AB=CD$?

It looks like it is supposed to be $AB=AD$ but I'm not entirely sure...

Quote: $AB=CD$? Sorry, that is $AB=AD$. And I have corrected it.

Edit: Forgive me I don't know what a fraction is

The only way that $C$ lies on the angle bisector of $\angle PAQ$ is if $\frac{EP}{PF} \cdot \frac{EQ}{QF} = \left( \frac{AE}{AF} \right)^2.$ I'm not completely sure how $\frac{AE}{EP}=\frac{AF}{FQ} \implies \frac{EP}{PF} \cdot \frac{EQ}{QF} = \left( \frac{AE}{AF} \right)^2$ (in fact, if this was true, we would have $EP = QF$ which would further imply $AE = AF!$) but then again, it's quite likely I might be missing something. Can you explain your black magic a bit more? Even though it seemed like (from the diagram WindyLi provided) that $P$ and $Q$ are isogonal conjugates, I don't think China Girls Math Olympiad 2020 would have gotten a problem wrong.

Inconsistent wrote: By symmetry, $C$ lies on the angle bisector of $\angle BAD$, and by ez fraction manipulation (black magic) using $EP = kAE, QF = kAF$, we know $C$ is also one the angle bisector of $\angle PAQ$. Extend $M = CP \cap AB$, $N = CQ \cap AD$, then it suffices to show $\frac{CP}{CM}=\frac{CQ}{CN}$ using that weird school geo lemma that no one likes, a.k.a. $CM*CX = CB^2$ by $\triangle CBX \sim \triangle CMB$, and same for other side together with pop on the desired quad. However, it was here that my LoS didn't work out I finally drew the diagram and GeoGebra and realized I had been lied to! Refund!!! 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If $C$ lies on the angle bisector of $\angle PAQ$, so $AE=AF$

Solution (scratch): Let CP intersect AB at E1 and CQ intersect AD at F1. Let <CBX = a, <CDY = b. Then, <BE1C = a, and <DF1C = b. CX = sin(a) * BC. CY = sin(b) * CD. So CX/CY = sin(a)/sin(b). We now wish to prove CP/CQ = sin(b)/sin(a). By LoS on CPQ, CP/CQ = sin(CQP)/sin(CPQ). By LoS on EE1P: EP/sin(a) = PE1/sin(AEF). By LoS on FF1Q: FQ/sin(b) = QF1/sin(AFE). Dividing the two equations we get: (EP/FQ) * (sin(b)/sin(a)) = (PE1/QF1) * (AE/AF). But EP/FQ = AE/AF so PE1/QF1 = sin(b)/sin(a). But also CB/CE1 = sin(a) and CD/CF1 = sin(b) so CE1/CF1 = sin(b)/sin(a). Thus, PE1/QF1 = CE1/CF1 so PE1/CE1 = QF1/CF1 and thus PQ is parallel to E1F1. So, CP/CQ = CE1/CF1 = sin(b)/sin(a) and thus CP/CQ = CY/CX. CX * CP = CY * CQ and thus PXYQ is cyclic.

Let,$CP\cap AB=M$ and $CQ\cap AD= N$.We have, $CX\times CM=CB^2=CD^2=CQ\times CN$.So it is enough to show by Reim's theorem that $MN||PQ$ $\textbf{\textcolor{red}{Claim:}}$$MN||PQ$. $\textbf{\textcolor{blue}{proof:}}$\begin{align} \frac{NC}{MC} =\frac{sin \angle BMC}{sin \angle CND}\\ =\frac{\frac{EP}{MP} sin\angle MEP}{\frac {QF}{QN}sin \angle NFQ}\\ =\frac{QN}{MP}\times \frac {EP}{QF}\times \frac{sin \angle E}{sin\angle F}\\ =\frac{QN}{MP}\times \frac {EP}{QF}\times \frac{AF}{AE}\\ =\frac{QN}{MP} \end{align}Where (1) is by right triangle $BMC$ and $NDC$ (2) is by sine law on $MEP$ and $QNF$ and (4) is by sine law on $AEF$. Hence, we are done by Reim's theorem on $X-P-M$ and $Y-Q-N$ and cyclic $XMNY$.$\blacksquare$

Mathematicsislovely wrote: We have, $CX\times CM=CB^2=CD^2=CQ\times CN$.So it is enough to show by Reim's theorem that $MN||PQ$ May I ask why can we multiply like this and where are the circles from? I can't see any concyclic ones... thanks

jchang0313 wrote: Mathematicsislovely wrote: We have, $CX\times CM=CB^2=CD^2=CQ\times CN$.So it is enough to show by Reim's theorem that $MN||PQ$ May I ask why can we multiply like this and where are the circles from? I can't see any concyclic ones... thanks Answer of your first question: Observe that,$BXC\sim BMC$ as $\angle MBC=\angle BXC=90^{\circ}$ and $\angle BCX$ is common in both triangle.So $\frac {CX}{CB}=\frac {BC}{CM}$.So $CB^2=CM\times CX$. Now observe that $CB^2= CD^2$.Now apply the similarity on $CND$ and $CDY$. Answer to your 2nd question: We got $\frac {CX}{CY}=\frac {CN}{CM}$ and note that $\angle XCY=\angle NCM$.So $CXY\sim CNM$.So $180^{\circ}-\angle MXY=\angle CXY=\angle CNM$.Hence, $XYNM$ is cyclic.

Thanks a ton!

Let P',E',X' be the reflections of P,E,X across AC, and M the intersection of AC and EF then: E'P'/FQ=AE/AF=EM/MF So QP' is parallel to AD Then a simple angle chase using parallels QP', AD and cyclic CDYX' yields cyclic QP'X'Y, So CY*CQ=CX'*CP'=CX*CP$

Mathematicsislovely wrote: Let,$CP\cap AB=M$ and $CQ\cap AD= N$.We have, $CX\times CM=CB^2=CD^2=CQ\times CN$.So it is enough to show by Reim's theorem that $MN||PQ$ $\textbf{\textcolor{red}{Claim:}}$$MN||PQ$. $\textbf{\textcolor{blue}{proof:}}$\begin{align} \frac{NC}{MC} =\frac{sin \angle BMC}{sin \angle CND}\\ =\frac{\frac{EP}{MP} sin\angle MEP}{\frac {QF}{QN}sin \angle NFQ}\\ =\frac{QN}{MP}\times \frac {EP}{QF}\times \frac{sin \angle E}{sin\angle F}\\ =\frac{QN}{MP}\times \frac {EP}{QF}\times \frac{AF}{AE}\\ =\frac{QN}{MP} \end{align}Where (1) is by right triangle $BMC$ and $NDC$ (2) is by sine law on $MEP$ and $QNF$ and (4) is by sine law on $AEF$. Hence, we are done by Reim's theorem on $X-P-M$ and $Y-Q-N$ and cyclic $XMNY$.$\blacksquare$ Don't you have to prove that $MNQP$ is cyclic?

Let K = BC x EF, L = DC x EF AI ⊥ EF( I lies on EF). PM ⊥ BC, QN ⊥ CD (M lies on BC, N lies on CD) (BM/KB)=(EP/KE), (KE/AE)=(KB/AI) => (BM/KB).(KB/AI)=(EP/KE).(KE/AE) =>(BM/AI)=(EP/AE) Also, (ND/AI)=(FQ/AF) Hence that BM=ND => CX.CP=CB.CM=CN.CD=CY.CQ (Q.E.D)(bacphoizbaoloc-27/10/2021)

LeHoangVu1208 wrote: Let K = BC x EF, L = DC x EF AI ⊥ EF( I lies on EF). PM ⊥ BC, QN ⊥ CD (M lies on BC, N lies on CD) (BM/KB)=(EP/KE), (KE/AE)=(KB/AI) => (BM/KB).(KB/AI)=(EP/KE).(KE/AE) =>(BM/AI)=(EP/AE) Also, (ND/AI)=(FQ/AF) Hence that BM=ND => CX.CP=CB.CM=CN.CD=CY.CQ (Q.E.D)(bacphoizbaoloc-27/10/2021)