Croatian Mathematical Society Competitions

a) Since $△DBC=△BAC$ and $△BCD=△BCA$, the triangles $BCD$ and $ABC$ are similar. Hence, $|BD|=|BC|$.

Let $E^{\prime }$ be the point such that the quadrilateral $A{E}^{\prime }BC$ is a parallelogram. Since $△AB{E}^{\prime }=△BAC=△CBD$, we get $△ABC=△CBD+△ABD=△AB{E}^{\prime }+△ABD=△E^{\prime }BD$. We also have $|B{E}^{\prime }|=|AC|$, so the triangles $E^{\prime }BD$ and $ABC$ are congruent. Therefore, the triangle $E^{\prime }BD$ is isosceles, and $E^{\prime }$ lies on the bisector of the segment $\overline{BD}$.

Since both $E$ and $E^{\prime }$ lie on the line passing through $A$ parallel to $BC$, we conclude that $E=E^{\prime }$. Hence, $AEBC$ is a parallelogram, i.e. the lines $AC$ and $BE$ are parallel.

b) Let $P$ be the midpoint of the segment $\overline{BC}$. Since (according to a)) the quadrilateral $AEBC$ is a parallelogram, we have $|BC|=|AE|$ and $|AE|:|CP|=2:1=|AF|:|CA|$.

The triangles $CAP$ and $AFE$ are similar, hence $△DFE=\frac{1}{2}△BAC$. We also have $△APC=90^{\circ }$ and $△AEF=90^{\circ }$.

Let $G$ and $H$ be the orthogonal projections of $F$ and $E$ to the lines $AB$ and $AC$, respectively. Let $T$ be the intersection of the lines $EH$ and $FG$. We must show that $T$ lies on the line $BD$.

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According to a), the triangles $EBD$ and $ABC$ are congruent, hence $$\angle ADE=180^{\circ }-\angle BDE-\angle BDC=180^{\circ }-\angle ACB-\angle ABC=\angle BAC.$$ Since $$\angle DFT=90^{\circ }-\angle FAG=90^{\circ }-\angle BAC=90^{\circ }-\angle ADE=\angle DEH=\angle DET,$$ we conclude that the quadrilateral $DEFT$ is cyclic, and $\angle DTE=\angle DFE=\frac{1}{2}\angle BAC$ holds. Finally, we have $\angle TDH=90^{\circ }-\frac{1}{2}\angle BAC$ and $\angle BDH=90^{\circ }+\frac{1}{2}\angle BAC$, hence $B$, $D$ and $T$ are collinear. This completes the proof.