This problem has been created by applying an inversion to an old problem. (See detailed solution i the attach file in iranian) Apply an inversion throw $P$ to get this problem: Problem: Let $P$ be a point on the arbitrary segment $AB$. choose arbitrary points $D$ and $C$ in the plain. Let $\omega$ with circumcenter $O$ be a circle which passes throw $C,D$ and tangent to $AB$ at $P$. $AD$ cuts $\omega$ again at $F$ and $BC$ cuts $\omega$ again at $E$. Let $AD\cap BC=W$. Let $X$ and $Y$ be the miquel points of $\{P,E,D\}$ and $\{P,C,F\}$ WRT $\triangle ABW$. then prove that: a) Prove that $O$ lies on $\odot (PXY)$. b) Prove that $OP$ bisects $\angle XPY$. Proof: General problem: $M,N,L$ are arbitrary points on sides $BC,CA,AB$ of $\triangle ABC$. $P$ is an arbitrary point on the plane of $\triangle ABC$ and $D,E,F$ are projections onto $BC,CA,AB.$ points $M',N',L'$ are the reflections of $M,N,L$ about $D,E,F$ and $U,V$ are the Miquel points of $MNL$ and $M'N',L'$ WRT $\triangle ABC.$ Then $PU=PV.$ this problem has been posted a lot of times: http://artofproblemsolving.com/community/c6h599366p3636067 (Generalization at post 6) http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=298400 http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=363074 (Lemma) From the special case of above generalization when $D\equiv M'\equiv M$ for triangle $WAB$ we immediately get $O$ is midpoint of arc $XY$ of $\odot (PXY)$ which proves the problem statements.

Attachments:
problem 6.pdf (47kb)

It's easy for p6! I'll write my solution briefly: a. It's enough to prove that $\triangle RCY \sim \triangle BXY$ $\triangle PFC \sim \triangle ABX \rightarrow \frac {(PC=RC)}{CF}= \frac {BX}{AB} = \frac {CR} {CY}$ $\rightarrow \frac {CR} {BX} = \frac {CY} {BY} (1)$ $AB || CD , \angle ABX = \angle PCD = \angle DCB \rightarrow \angle XBY = \angle RCY , (1) $ $\rightarrow \triangle XBY \sim \triangle CRY \rightarrow \angle BYX = \angle CYX \blacksquare$ b.$\angle AFY = \angle PDX (*)$ $\triangle CFY \sim \triangle PED , \triangle PFC \sim \triangle DEX$ $\rightarrow \frac {PF} {DX} = \frac {CF} {DE} = \frac {CY} {PE} = \frac {FY} {PD} , (*) \rightarrow \triangle PFY \sim \triangle PDX$ $\rightarrow \frac {PY} {PX} = \frac {CF} {DE} =\frac {\frac {CY} {BY}} {\frac {XD} {AX}} =\frac {\frac {RY} {XY}} {\frac {RX} {XY}} = \frac {RY} {RX}$ $\rightarrow \angle XPR = \angle RPY \blacksquare$

Another extension: Two circles $\omega_1,\omega_2$ intersect at $P,Q $. An arbitrary line passing through $P $ intersects $\omega_1 , \omega_2$ at $A,B $ respectively. choose points $F,D$ on $\omega_1$ and $E,C$ on $\omega_2$. assume there exist a point $R$ such that: $$\frac{PD}{PF}=\frac{RD}{RF}\ ,\frac{PC}{PE}=\frac{RC}{RE}\ ,RP\perp AB$$a) prove that $R$ lies on $XY$. b) prove that $PR$ is bisector of $\angle XPY$.

PS

Attachments:

$\angle XDQ = \angle APQ = \angle BCQ = \angle XEQ$. Thus $XDEQ$ is cyclic. Similarly $YCFQ$ is cyclic. Let $S$ be the point of intersection of $(XDEQ)$ and $(YCFQ)$. Notice: \begin{align*} \angle XSQ &= \angle XEQ \\ &= \angle BCQ \\ &= 180 - \angle QCY \\ &= 180 - \angle QSY, \end{align*}thus $S$ lies on the line $XY$. Also: \begin{align*} \angle ESF &= 180 - \angle XSE - \angle FSY \\ &= 180 – \angle ADE - \angle ACF \\ &= 180 - \angle FPC - \angle APE \\ &= \angle EPF \\ &= \angle ERF, \end{align*}thus $EFRS$ is cyclic. Finally: \begin{align*} \angle FSR &= \angle FER \\ &= \angle PEF \\ &= \angle BCF \\ &= \angle YSF, \end{align*}which implies that $R$ lies on the line $SY$ which let us conclude (a). As we pointed out before $\angle FER = \angle BCF$, thus $ER \parallel BC$, similarly $FR \parallel AD$. Then $\tfrac{AF}{FY} = \tfrac{XR}{RY} = \tfrac{XE}{EB} = \tfrac{XD}{DA}$. Also notice that $\angle XDP = \angle YFP$ and for our last observation: $\tfrac{YF}{DP} = \tfrac{YF}{FA} = \tfrac{AD}{DX} = \tfrac{PF}{DX}$. Thus by SAS, $\triangle PYF \sim \triangle XPD$. Finally: $\tfrac{PY}{PX} = \tfrac{PF}{XD} = \tfrac{YR}{RX}$, and for the angle bisector theorem, we conclude part (b).

Solution. By construction of $R$, $\angle DFR=\angle DFP=\angle ADF$, so $FR\parallel AD$. Further, $FR=FP=AD$, so $ADRF$ is a paralelogram. In a similar fashion, we can prove that $BCRE$ is a paralelogram as well. Let $Y'=\overline{XR}\cap\overline{AF}$. Note that $\frac{XE}{XB}=\frac{XD}{XA}=\frac{XR}{XY'}$, thus $BY'\parallel ER$, but $BY\parallel ER$, hence $B,\ Y,\ Y'$ are collinear, which implies $Y=Y'$. Hence, $R$ lies on $XY$. For the second part, construct the points $Z=\overline{CD}\cap\overline{XP}$ and $T=\overline{CD}\cap\overline{YP}$. We have $\frac{XZ}{XP}=\frac{XD}{XA}=\frac{XR}{XZ}$, so $ZR\parallel PY$. Analogously, $TR\parallel PX$. Since $PR\perp CD$, we infer that $PZRT$ is a rhombus, whence $\angle XPR=\angle ZPR=\angle RPT=\angle RPY$, as required.

Asymptote please.

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Diagram

$ADRF$ and $BERC$ are parallelogram. Let $DR'$ || $AY$ where $R'$ KORS in $XY$. Then $XR'/XY=XD/XA=XE/XB$ => $ER' || BY$ or $R=R'$. Let $K=(DA \cap CD)$, $L=(AY \cap BX)$, $T=(AB \cap XY)$. Now, $\triangle REF$ and $\triangle KBA$ are homothetic and similar. So $K,L,R$ are collinear. Implies $(X,Y;R,T)=-1$. $PR \perp AB$ so $PR$ bisects $\angle XPY$.

a) Note that by some Easy angle chasing we have triangles BAX and CFP are similar. R is reflection of P across DC so BAX and CFR are similar also CF || BA and BX || CR and AX || FR so by homothety we have BC,AF,XR are collinear at a point which is Y so Y lies on XR. b) PX/PY = PD/FY = DE/CF = DE/AB / CF/AB = XD/XA / YF/YA = XR/XY / YR/YX = XR/YR ---> PR is angle bisector of XPY.