Clearly $S$ is the center of spiral similarity from $BA$ to $DC$, so we have $BSA\sim CSD$ and $ASC\sim BSD$. Consider a point $L$ on the tangent line to $(ASD)$ at $S$, such that $\angle LSA<\angle LSD$. We have \[\angle LSB=\angle ASB-\angle LSA=\angle CSD-\angle SDA=\angle CSD-(\angle ADC-\angle SDC)=\]\[\angle CSD+\angle SDC-(180^{\circ}-\angle BCD)=\angle BCD-\angle SCD=\angle SCB\]which shows that $SL$ is tangent to $(BSC)$ as well, as desired.

Let the tangent to $(BSC)$ cut $AB$ at $E$. Let $\angle ESA = \alpha , \angle BSE = \theta , \angle BCA = \gamma$. We have $\alpha + \theta = \angle BSA = \angle BOA \implies \angle DBC = \alpha + \theta - \gamma = \angle BDA$. Finally note that because of the tangency, $\theta = \angle BCS \implies \angle ODS = \angle OCS = \theta - \gamma$. And so $\angle SDA = \angle BDA - \angle BDS = \alpha = \angle ESA$, implying the result. $\square$

Attachments:

It is enough to prove $\angle ADS+ \angle SCB=\angle ASB$. $\angle ADS+\angle SCB= \angle OBC+\angle OCS+\angle SCB=\angle AOB=\angle ASB$ as desired. $\square$

First $\angle ABS=\angle AOS=180-\angle SOC=\angle SDC$ and with a similar way, $\angle SAB=\angle SCD$, hence $\angle ASB=\angle DSC$. Let $X$ be the intersection of $AD$ and the tangent line of $ASD$ in $S$ and $M$ the intersection of $AD$ and $SB$. Then:$\angle XSC=\angle XSD+\angle DSC=\angle SAD+\angle ASB=\angle AMB=\angle MBC=\angle SBC$. Therefore, $SX$ is also tangent to $BSD$ and we are done.

Note that we only need to prove $\angle SBC + \angle SAD = \angle DSC$. $\angle DSC = \angle DOS = \angle OBC + \angle OCB = \angle OAD + \angle SBC + \angle OBS = \angle DAS + \angle SBC$. we're Done.

I took part to this contest last year,it was a pretty interesting problem. I solved it whith a little bit different method compare to the officiel solution.sorry for my bad english

We have $\measuredangle ASB=\measuredangle AOB=\measuredangle COD=\measuredangle CSD$ and $\measuredangle SDC=\measuredangle SOC=\measuredangle SOA=\measuredangle SBA$ so $\triangle SDC\sim \triangle SBA$ Now let $\ell$ be the tangent of $(SAD)$ through $K$, and let $K$ be a point on it. We have $\measuredangle KSD=\measuredangle SAD=\measuredangle SAB+\measuredangle BAD=\measuredangle SCD+\measuredangle ABC$ and $\measuredangle KSC=\measuredangle KSD+\measuredangle DSC=\measuredangle SCD+\measuredangle ABC+\measuredangle DSC=\measuredangle SDC+\measuredangle ABC=\measuredangle SBA+\measuredangle ABC=\measuredangle SBC$ So $KS$ is tangent to $(SBC)$.

anyone__42 wrote: Let $ABCD$ be a trapezoid which is not a parallelogram, such that $AD$ is parallel to $BC$. Let $O=BD\cap AC$ and $S$ be the second intersection of the circumcircles of triangles $AOB$ and $DOC$. Prove that the circumcircles of triangles $ASD$ and $BSC$ are tangent.

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