BMO 2019 Shortlist

We have $$\{\begin{array}{l}DC\equiv DA\\ \angle EDC\equiv \angle EDA\\ DE\equiv DE\end{array}\Rightarrow △DEC\equiv △DEA\Rightarrow \angle DAE\equiv \angle DCE(\ast ).$$ Let $CM\cap AD=\{P\}$, then follows $△CDP\equiv △BAP$ and $\angle PCD\equiv \angle PBA(\ast \ast )$.

Figure 1: G1

From $(\ast )$ and $(\ast \ast \ast )$ follows $\angle DCP\equiv \angle DAE\equiv \angle PBA$.

Now, let $S^{\prime }=AE\cap PB$. In the triangle $S^{\prime }AB$ we have $$m(\angle S^{\prime }AB)+m(\angle S^{\prime }BA)=m(\angle S^{\prime }AB)+m(\angle PA{S}^{\prime })=m(\angle PAB)=90^{\circ },$$ so $m(\angle B{S}^{\prime }A)=90^{\circ }$.

We show that $AE$, $BP$ and $MO$ are concurrent. In the triangle $△EMB$ we apply the Ceva theorem, so $$\frac{EP}{PM}\cdot \frac{MA}{AB}\cdot \frac{BO}{OE}=1\iff \frac{EP}{PM}=\frac{OE}{BO}$$ is true because $PO$ is a midsegment in the triangle $DAB$ ($PO\parallel AB$).

According to the Thales theorem in the triangle $EMB$, $\frac{EP}{PM}=\frac{EO}{OB}$ and $AE$, $BP$, $MO$ are concurrent in $S^{\prime }$, which is in fact $S$.

Let $PB\cap CA=\{N\}$. Because $ESNO$ has $m(\angle EON)+m(\angle ESN)=180^{\circ }$, it follows $ESNO$ cyclic and $m(\angle ESO)=m(\angle ENO)=m(\angle DAO)=45^{\circ }$. □