China Girls' Mathematical Olympiad

Solution 1. Construct point $F_1$ such that $E{F}_1=B{F}_1$ and ray $D{F}_1$ is perpendicular to line $ED$. It suffices to show that $F=F_1$ or $ABD{F}_1$ is cyclic; that is, $$\angle BAD=\angle B{F}_{1}D.\stackrel{◯}{1}$$ Set $\angle BAD=\angle CAD=x$. Because $EC\perp AB$ and $AD\perp BC$, $$\angle ECB=90^{\circ }-\angle ABD=\angle BAD=x.$$ Note that $MD$ is a midline of triangle $BCE$. In particular, $MD\parallel EC$ and $$\angle MDB=\angle ECD=x.\stackrel{◯}{2}$$ In isosceles triangle $E{F}_{1}M$, we may set $\angle E{F}_{1}M=\angle B{F}_{1}M=y$. Because $EM\perp M{F}_1$ and $MD\perp D{F}_1$, $$\angle EM{F}_1=\angle ED{F}_1=90^{\circ },$$ implying that $EMD{F}_1$ is cyclic. Consequently, we have $$\angle EDM=\angle E{F}_{1}M=y.\stackrel{◯}{3}$$ Combining ② and ③, we obtain Fig. 2. 2 $$\angle BDE=\angle EDM+\angle MDB=x+y.$$ Because $BE=BD$, we conclude that triangle $BED$ is isosceles with $\angle MED=\angle BED=\angle BDE=x+y$. Because $EMD{F}_1$ is cyclic, we have $\angle M{F}_{1}D=\angle MED=x+y$. It is then clear that $$\angle B{F}_{1}D=\angle M{F}_{1}D-\angle M{F}_{1}B=x=\angle BAD,$$ which is ①.

Solution 2. (Based on work by Sherry Gong and Inna Zakharevich) We maintain the notations used in Solution 1. Let $\omega$ and $O$ denote the circumcircle and the circumcenter of triangle $ABD$, and let $T$ be second intersection (other than $B$) between line $BE$ and $\omega$. Extend segment $DE$ through $E$ to meet $\omega$ at $S$. Point $F_2$ lies on $\omega$ such that $D{F}_2\perp DE$. We will show that $F_2=F$ or $F_{2}B=F_{2}E$. Let $M_2$ denote the foot of the perpendicular from $F_2$ to line $BE$. It suffices to show that $M_2$ is the midpoint of segment $BE$, that is, $E{M}_2=M_{2}B$. Because $\angle SD{F}_2=\angle ED{F}_2=90^{\circ }$, $O$ is the midpoint of $S{F}_2$. Because $BTAD$ is cyclic, $\angle BTD=\angle BAD=\angle BCE=x$. Note also that $BD=BE$ and $\angle EBC=\angle DBT$. We conclude that triangle $BDT$ and $BEC$ are congruent to each other, implying that $BE=ET$. Hence, $OE\perp BT$. Let $N$ be the foot of the perpendicular from $S$ to line $BE$. Fig. 2. 3 Note that segments $NE$ and $E{M}_2$ are the respective perpendicular projections of segments $SO$ and $O{F}_2$ onto line $BE$. Because $SO=O{F}_2$, $NE=E{M}_2$. Because $TE=EB$, it suffices to show that $TN=NE$, which is evident since $$\angle STE=\angle SEB=\angle SDB=\angle EDB=\angle BED=\angle SET$$ and so triangle $SET$ is isosceles with $SE=ST$.