Assume the circumcircles of $ABY$ and $AEX$ meet at $A$ and $K$. $\angle KEX=\angle KAX=180^\circ-\angle KAY=180^\circ-\angle KBY=\angle KBS$. So, $(KBES)$ is concyclic. $\angle YKB=\angle YAB=\angle BCA=\angle PYB$ $\angle EKX=\angle EAX=\angle ADE=\angle QXE$ In solution, $\angle YZX=\angle YSX-\angle SYZ-\angle SXZ=180^\circ-(\angle BKE+\angle YKB+\angle EKX)=180^\circ-\angle YKX$ So, $(YKXZ)$ is concyclic. Let this circle $\Omega$. Let's say the tangent of $\Omega$ at $K$ meet $XY$ at $T$. $\angle TKB=\angle TKY+\angle YKB=\angle KXT+\angle YAB=\angle KEA+\angle AEB=\angle KEB$ As a result, $TK$ is a tangent of the circumcircle of $KEB$. So, the to circles are tangent at $T$.

Kind of easy for #3. Invert in $A$. Then $\overline{A-X-Y}\parallel\overline{B-C-D-E-F}.$ $Q$ becomes the second intersection of $AD$ with $(XED)$, $P$ becomes the second intersection of $AC$ and $(YBC)$, while $Z$ becomes the second intersection of $(XAQ)$ and $(AYP)$ and $S$ becomes the second intersection of $(AEX)$ and $(ABY).$ Take $T=XE\cap YB.$ All angles are oriented. Note that $\angle{XZA}=\angle{XQA}=\angle{XQD}=\angle{XEB}$ and in the same way $\angle{YZA}=\angle{YBE}$, so we can easily see that $T\in (XYZ)$. Also note that $\angle{ESA}=\angle{EXA}=\angle{XEB}$ and in the same way $\angle{BSA}=\angle{YBE}$, so we can easily see that $T\in (BES).$ The porblem thus reduces to showing that $(TXY)$ and $(TBE)$ are tangent, which is clear since $XY\parallel BE.$

Nice problem.We'll prove a more general problem that is we'll prove this problem for any position of $X,Y$.

At first I was like " A pentagon with lot of points on it and a tangency from nowhere?!" but I've got better when it turns $C,D$ are actually unnecessary; In fact , by changing $C,D$ over $(ABCDE)$ and fixing $X,Y,A,B,E$ , not only the tangency point but the whole $(XYZ) , (BES)$ remain unchanged... We have $\angle EXZ = \angle EDA = \angle EAX$ . Similarly , $\angle BYZ = \angle BAY$ .So $XZ,YZ$ are tangent to $(AXE),(AYB)$ respectively. Let $T = (AYB) \cap (AXE)$ . Now $\angle TXZ + \angle TYZ = \angle TAX + \angle TAY = \pi$ and $TXYZ$ is cyclic. Also , $\angle BTE = \angle BTA + \angle ATE =\angle AYS+\angle AXS = \pi - \angle BSE$ which implies that $TBSE$ is cyclic. So $T = (BSE) \cap (XYZ)$. Now the angle between $TX$ and the tangent through $T$ to $(TXZY)$ is $\angle TYX = \angle TBA$ and the angle between $TB$ and the tangent through $T$ to $(TBSE)$ is $\angle TEB = \angle TEA + \angle AEB$. Now since $\angle TXY + \angle TYX = \angle XZY$ we have the difference between the recent values is in fact $\angle XTB$ which implies that the tangents through $T$ are equal lines and we're done.

Claim $: ZY,ZX$ are tangent to $ABY$ and $AEX$. Proof $:$ Note that $\angle ZYS = \angle PYS = \angle PCB = \angle ACB = \angle YAB \implies ZY$ is tangent to $ABY$. we prove the other one with same approach. Note that $\angle BSE = \angle 180 - (\angle 180 - \angle AYB + \angle 180 - \angle AEX)$ and $\angle YZX = \angle 180 - (\angle ZYX + \angle ZXY) = \angle 180 - (\angle YBA + \angle XEA)$ so if $AEX$ and $ABX$ meet at $R$ we have $BSER$ and $XZYR$ are cyclic so Now we need to show they are tangent at $R$. Let $N$ be arbitrary point such that $RN$ is tangent to $XZYR$, we have $\angle ERN = \angle XRN - \angle XRE = \angle XYR - \angle XAE = \angle AYR - \angle EBA = \angle ABR - \angle ABE = \angle EBR \implies ESBR$ is tangent to $RN$ at $R$. One can also prove that $S,Z,R$ are collinear.

Dadgarnia wrote: Given an inscribed pentagon $ABCDE$ with circumcircle $\Gamma$. Line $\ell$ passes through vertex $A$ and is tangent to $\Gamma$. Points $X,Y$ lie on $\ell$ so that $A$ lies between $X$ and $Y$. Circumcircle of triangle $XED$ intersects segment $AD$ at $Q$ and circumcircle of triangle $YBC$ intersects segment $AC$ at $P$. Lines $XE,YB$ intersects each other at $S$ and lines $XQ, Y P$ at $Z$. Prove that circumcircle of triangles $XY Z$ and $BES$ are tangent. Let $X'$, $Y'$ in order are intersection of lines $AB$, $AE$ with circle $(SBE)$ ($X' \neq B$, $Y' \neq C$) First, we'll prove that there exist a homothety centered taking $\triangle X'SY'$ to $\triangle XZY$ : $\, \,$Cuz $XY$ is tangent to $\Gamma$ at $A$ so $(XY, AB) = (EA, EB) = (X'Y', X'B)$ (mod $\pi$). Lead to $X'Y'$ $\|$ $XY$ $\, \,$Cuz $Y$, $P$, $B$, $C$ are concyclic, so $(YZ, YB) = (CP, CB) = (EA, EB) = (SY', SB)$ (mod $\pi$). Lead to $ZY$ $\|$ $SY'$. Similar, $ZX$ $\|$ $SX'$ $\, \,$ Therefore, $\triangle SX'Y'$ and $\triangle ZXY$ are similar and oriented in the same way. $\, \,$Which means there exist a homothety centered taking $\triangle X'SY'$ to $\triangle XZY$ Thus, If intersection $T$ of lines $XX'$ and $YY'$ lies on circle $(SBE)$, we ez to see that cirles $(SBE)$, $(ZXY)$ are tangent at $T$ Hence, we just need to prove that intersection of lines $XX'$ and $YY'$ lies on circle $(SBE)$ : $\, \,$Let $T$ is the intersection of line $XX'$ and circle $(SBE)$ $\, \,$Use Pascal theorem for $\binom{S\ X'\ Y'}{T\ E\ B}$, we see that $Y'$, $Y$, $T$ are colinear (Cuz $Y$ is the intersection of lines $AX$, $SB$) Done.[/quote]