In the figure below we have $AX = BY$ . Prove that $ \angle XDA = \angle CDY$ . Proposed by Iman Maghsoudi

$YCD =45$ since $AY=YD$ let $YDC =x$ let $XDA=a$ let $BY=BC=AX =s$ $YC = 2s\cos(x)$(Sine Rule) $YD = \frac{s\cos(x)\sqrt{2}}{\sin(x)}$(Sine Rule) since $\frac{SD}{\sin(45)} = \frac{2s\cos(x)}{\sin(x)}$ Sine Rule Or Pythag Suffice thus $AD= \frac{s\cos(x)}{\sin(x)}$ $\frac{s}{\sin(a)} =\frac{\frac{s\cos(x)}{\sin(x)}}{\cos(a)}$ tan(a) = tan(x) , $a=x$

If $O$ is the center of the circle, then $AYOD$ is a square and $\triangle BYO$ is congruent to $\triangle XAD$. Then $\angle CDY = \frac{1}{2} \widehat{CY} = \angle BOY = \angle XDA$.

Let $O$ be the center of the circle. $ADOY$ is a square, so $DO = AY = AX+XY = BY+XY = XB$. Also $XB||DO$, so $BXDO$ is a parallelogram. Now $\angle XDA = 90 - \angle AXD = \angle YXD - 90 = 90 - \angle YBD = \angle YOB = \frac{\angle YOC}{2} = \angle CDY$.

Let: $O$ circumcenter of $\omega$ So $ OY\perp AB, OC\perp CB$ \[\angle YOC=\angle CYB= \angle YOB = \beta\]But $OY=DA, AX=YB, \text{ and } \angle DAX= \angle OYB = 90^{\circ}$ So $\triangle DAY\cong \triangle OYB$ \[\angle ADX= \angle YOB = \angle CDY = \angle XDA \blacksquare\] As desired

Let $O$ be the center of $\omega$. We have, $AX=BY$, $\angle DAX=\angle OYB$ and $AD=OY$. Which implies, $\Delta ADX\cong\Delta BOY\implies\angle ADX=\angle YOB=\angle BOC\implies\angle ADX=\frac{1}{2}\angle COY=\angle CDY$

Let $O$ be the circumcenter of $w$. Then, $\angle ADX=\angle YOB=\frac{1}{2}\angle YOC=\angle YDC$.