Let $CD$ intersect $w$ again at $W$. Now Pascal's on degenerate hexagon $WXYBCC$ yields that the tangent to $w$ at $C$ and $XY$ intersect on $AD$, implying that $Z$ is on $XY$ as desired.

Non-projective geometry solution: Let $XY$ intersect $AD$ at $Z'$. We will show that $Z' \equiv Z$. Let $XY$ intersect $BC$ at $M$. Call angle $YCM$ $\alpha$, and $YMC$ $\beta$. We get that $Z'YC=\alpha + \beta$, and $ABC= \alpha +\beta$ ($BXY=BCY$, $BMX=YMC$). From the second one, with the conditions, $DCB= \alpha + \beta$. But since $YCB$ is $\alpha$, $YCD=\beta$. This is great, because trivially $BMX=AZ'X=\beta$, meaning that $DZ'Y=180-\beta$ and $DCYZ'$ is a cyclic quadrilateral. We are almost done, note that from this, $YZ'C=YDC$, and with that, triangles share two angles in common ($YZ'C=YDC$, and $Z'YC=DCB$ because $MYC=180-(\alpha +\beta)$, so their third angles are equal too, $Z'CY=DBC=YBC$. But because the chord angle of $YC$ is $YBC$ (in $\omega$), $YCZ'$ is a tangent chord angle, so $CZ'$ is a tangent and we are done, $Z' \equiv Z$.

Delray wrote: Let $CD$ intersect $w$ again at $W$. Now Pascal's on degenerate hexagon $WXYBCC$ yields that the tangent to $w$ at $C$ and $XY$ intersect on $AD$, implying that $Z$ is on $XY$ as desired. I don't clearly understand this proof, because I am not sure what the Pascal Theorem is like on degenerate hexagons. I'll mark my questions. 1)How can even you use the Pascal theorem on degenerate hexagon? Like, where do I find a proof of it working if you use a point more than once.. 2) If so, then from my understanding, the 6th side (which is CC) is a tangent in C? 3)How can you even choose the sequence of points to be WXYBCC, when clearly B is between X and Y? Was that a typo, or you can mark the points with 1,2,3,4,5,6 in any way you want? Please if yes, someone send me a proof/a good paper about Pascal's and Projective Geometry... 4) If, assuming its WXYBCC, then you're searching for the intersection point of WX and CD which clearly doesn't exist.. Then what does it "become" in the theorem? Does the theorem state that the other two points' segment is paralel to CD and WX? Just please give some answers to my questions..

When Pascal's theorem is used on a degenerate hexagon, then $AA$ denotes the tangent passing through $A$. To answer question 3), Pascal's does not have to have the points chosen in order, as long as they are 6 points on a conic section. For question 4, as Pascal's is on the projective plane, WX and CD intersect at a point at infinity, implying that the other line passes through that point at infinity too.

Thank you, I almost understand everything after ut after this, I still kinda dont get a conclusion from question 4's answer: okay I get it now but how does that mean that we are done? Like.. I think that somehow gives us that the line that connects D with point of infinity is paralel with XW and CD, but can't explain totally why.

Proof by Angle chasing Step 1 Let $XY$ meet $AD$ at $Z'$. Step 2 By angle chasing, prove that $CYDZ'$ is cyclic. By this we get that $\angle CXY=\angle YCZ'$ hence $Z'C$ is tangent so $Z'=Z$ as desired. EDIT- I just realized it has been posted above..

Suppose $CD$ meets $\omega$ again at $W$. We have $\angle ADY=\angle ADB=\angle CBY=\angle ZCY\implies$ $CYDZ$ is cyclic. Thus \[\angle BYX = \angle BCX=\angle CBW=\angle ZCW=\angle ZYD.\]Proved.

Let $CD$ meet $\omega$ H. $\angle ZCY=\angle YBC=\angle DBC=\angle ADB\implies$ $CYDZ$ is cyclic. It is easy to prove that $XH$$\parallel$$CD$ so, $BXHC$ is isosceles trapezoid . $\angle DCZ=\angle HBC=\angle XCB=\angle XYB\implies$ $X$,$Y$,$Z$ are collinear

Here's my solution: Let $CD \cap \omega=T$. Then $BXTC$ is an isosceles trapezoid, which gives that $\angle ZCD=\angle CXT=\angle XCB$. This means that $CB,CZ$ are isogonal in $\angle DCX$. Taking $XZ \cap BD=Y'$, and applying Isogonal Line Lemma, we get that $CY'$ and $CA$ are also isogonal in $\angle DCX$. Thus, $$\angle XCY'=\angle DCA=\angle DBA=\angle XBY' \Rightarrow Y' \equiv Y \quad \blacksquare$$

Let $T$ be the point of intersection of line $CD$ with line parallel to $AD$ passing through $X$. Since $X$ lies on segment $AB$, point $T$ lies on segment $CD$. Trapezoid $ABCD$ is isosceles thus $$\angle BXT=\angle BAD=180^\circ-\angle BCT$$which means quadrilateral $BXTC$ is cyclic: \angle TBC=\angle BTX=\angle BYX There exists exactly one circle circumscribed on triangle $BCX$ which means points $B,X,Y,T,C$ are concyclic. The tangent to $\omega$ at $C$ intersects the line $AD$ at $Z$: $$\angle TBC=\angle DCZ$$But $$AD||BC\implies \angle CDZ=\angle BCT$$so $$\angle DZC=\angle BTC=\angle BYC=180^\circ-\angle DYC$$. Hence quadrilateral $CYDZ$ is cyclic and $$\angle DYZ=\angle DCZ=\angle TBC=\angle BYX$$Points $B,Y,D$ are collinear so equality $\angle DYZ=\angle BYX$ gives collinearity of points $X,Y,Z$.

First, notice that $\angle BDA=\angle DBC=\angle YBC=\angle YCZ$. Therefore, $CZDY$ is cyclic. Also, $\angle CDA= \pi-\angle DCB=\pi-\angle CBA=\pi-\angle CDZ \implies \angle CBA=\angle CDZ$. By $AA$ triangle similarity, $\triangle CBX ~ \triangle CDZ \implies \angle BYX=\angle BCX= \angle DCZ= \angle DYZ$. Therefore, $X, Y, Z$ are collinear.

utkarshgupta wrote: $ABCD$ is an isosceles trapezoid with $BC || AD$. A circle $\omega$ passing through $B$ and $C$ intersects the side $AB$ and the diagonal $BD$ at points $X$ and $Y$ respectively. Tangent to $\omega$ at $C$ intersects the line $AD$ at $Z$. Prove that the points $X$, $Y$, and $Z$ are collinear. Let $\omega \cap CD = T$. Then we have that $$\measuredangle DTX = \measuredangle CTX = \measuredangle BCA \implies TX \parallel AD$$Now finish of with Pascals theorem on $CCBYXT$

Let $\angle XYB=\alpha ; \angle ZYD=\beta$ $\implies \angle XYB=\angle XCB=\alpha$ Let $\angle YBC=\gamma \implies \angle YCD=\gamma-\alpha$ but $\angle YBC=\gamma=\angle YCZ \implies \angle DCZ=\alpha$ As well: $\angle ADB=\angle DBC=\angle YCZ=\gamma \implies DYCZ$ is cyclic. $\implies \angle DCZ=\angle DYZ \implies \alpha=\beta$ $\implies X,Y,Z$ are collinear. $\blacksquare$

Define $Z’=AD \cap XY$ Now just notice that a spiral sim centred at $C$ Takes $AD \longrightarrow XY$ Which means $C$ Is the Miquel point of $AXYD$. This means $\odot(ZDYC)$ so $$\angle YBC = \angle DBC = \angle BDA = \angle Z’CY \implies Z=Z’ \implies X-Y-Z \ \blacksquare$$

$\textbf{\textcolor{red}{Claim 1:-}}$ $CZDY$ is a cyclic quadrialteral. $\textbf{\textcolor{blue}{Pf:-}}$ Since $BC \parallel AD$ we have $\angle{CBY}=\angle{ADB}$ and since $CZ$ is tangent we have $\angle{CBY}=\angle{ZCY}$. Since $\angle{YDZ}=180^{\circ}-\angle{ADB}$ we have $CZDY$ to be cyclic quadrialteral. $\square$ Denote $W:= CD \cap \omega$ $\textbf{\textcolor{red}{Claim 2:-}}$ $WCXB$ is a isosceles trapezoid. $\textbf{\textcolor{blue}{Pf:-}}$ Join $WX$. Since $WCXB$ is cyclic we have $\angle{XWC}=180^{\circ}-\angle{CBX}$ , so $\angle{DBX}=\angle{CBX}$ and since $BC \parallel AD$ we have $\angle{ADC}=180^{\circ}-\angle{CBX}$ Hence $AD \parallel WX \parallel CB$ Since $WCXB$ is cyclic and $WX \parallel CB$ we get $WCXB$ as isosceles trapezoid. $\square$ Since $\angle{XYB}=\angle{XWB}=\angle{XCB}$ and also , $\angle{XYB}=\angle{WBC}$ as $CZ$ is tangent we have $\angle{WBC}=\angle{ZCD}$, hence $\angle{XYB}=\angle{ZYD}=\angle{ZCD}$ , Hence $X,Y,Z$ has to be collinear. $\blacksquare$