Croatian Mathematical Society Competitions

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

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

Since both $E$ and $E'$ lie on the line passing through $A$ parallel to $BC$, we conclude that $E = E'$. 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 $\triangle DFE = \frac{1}{2}\triangle BAC$. We also have $\triangle APC = 90^\circ$ and $\triangle AEF = 90^\circ$.

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.