Solution:

For convenience of notation, let $\overline{BAC} = \alpha$ and $\widehat{ABC} = \widehat{ACB} = \beta$.

Let us draw the circle circumscribed about triangle $BDA$ and let $E'$ be its intersection with the parallel to $BC$ through $A$; we will show that $E'$ lies on the perpendicular bisector of $BD$, and must therefore coincide with $E$. We have $\widehat{BAD} = \widehat{BE'D} = \alpha$ (both subtend $BD$), $\widehat{E'DB} = \widehat{E'AB} = \widehat{ABC} = \beta$ (the last equality follows from the parallelism of lines $E'A$ and $BC$); hence triangle $BE'D$ is similar to triangle $BAC$, and is therefore isosceles. It follows that $E'$ lies on the perpendicular bisector of the base $BD$,

as required. Moreover we have thus obtained that $\widehat{BEA}$ must be supplementary to $\widehat{ADB}$, that is, equal to $\beta$. This proves the parallelism between line $EB$ and line $AC$.

Now let $P$ be the point of intersection between lines $r$ and $s$, and $Q$ the point of intersection between lines $r$ and $t$; call $H$ the projection of $F$ onto line $AB$. By construction, line $r$ is perpendicular to line $AB$, while line $s$ is perpendicular to $AC$, hence to its parallel $BE$; in other words, angles $\widehat{BEP}$ and $\widehat{BHP}$ are right angles, so quadrilateral $BEHP$ is cyclic. It would suffice to show that $Q$ lies on the circle circumscribed about $BEHP$ to conclude that $P$ and $Q$ coincide (since both are points on line $r$ different from $H$), from which the concurrency of lines $r$, $s$ and $t$ follows.

The angle $\widehat{BQH}$, since $\widehat{BHQ}$ is a right angle, equals $90^\circ - (\widehat{ABC} - \widehat{CBD}) = 90^\circ + \alpha - \beta$. Note that quadrilateral $AEFH$ is cyclic: $\widehat{AHF}$ is a right angle by construction; $EA$ is congruent to $BC$, $AF$ is twice $AC$, $\widehat{EAF} = \beta$, and hence triangle $EAF$ is similar to the triangle formed by $A$, $B$ and the foot of the altitude of $ABC$ from $A$, so the angle $\widehat{FEA}$ is also a right angle. Observing that both $\widehat{FEH}$ and $\widehat{FAH}$ subtend the arc $FH$, we have shown $\widehat{FEH} = \widehat{FAH} = \alpha$. We consequently obtain $\widehat{BEH} = \widehat{BEA} + \widehat{AEF} - \widehat{FEH} = \beta + 90^\circ - \alpha$; that is, angles $\widehat{BQH}$ and $\widehat{BEH}$ are supplementary, so quadrilateral $BQHE$ is cyclic. We have thus shown that $Q$ belongs to the circle circumscribed about $BEHP$, whence the claim.