, 2008

Yes. Draw $AK$ parallel to $BC$ and $AK = BD$. Join $K$ to $E$ and $F$. Choose $L$ on $BC$ with $L$ between $B$ and $C$, such that $BD = LC$. Note that $KB$ is parallel to $AD$ and $KB = AD$. Hence $\angle KBE = \angle APE = 60^\circ$. On the other hand, $KB = AD = BE$. Thus $KBE$ is an equilateral triangle. This gives $\angle BEK = 60^\circ = \angle BQR$. Thus $CF$ is parallel to $EK$ but $CF = BE = EK$. We conclude that $FCEK$ is a parallelogram. Since $CL$ is parallel to $AK$ and $CF$ is parallel to $KE$ we have $\angle LCF = \angle AKE$.

Consider the triangles $LCF$ and $AKE$. We have $LC = AK$, $CF = KE$ and the included angles are also the same. Hence $LCF$ is congruent to $AKE$. Moreover $LF$ is parallel to $CA$. Thus
$$\frac{BF}{FA} = \frac{BL}{LC} = \frac{CD}{BD} = \frac{1}{\lambda}, \text{ say.}$$
Consider the transversal $FQC$ of the triangle $ABD$. By Menelaus' theorem
$$\frac{BF}{FA} \cdot \frac{AP}{PD} \cdot \frac{CD}{CB} = 1.$$
Thus
$$\frac{AP}{PD} = \frac{FA}{BF} \cdot \frac{CB}{CD} = \lambda \left( \frac{CD + DB}{CD} \right) = \lambda(1+\lambda).$$
Similarly, we get
$$\frac{BQ}{QE} = \frac{CR}{RF} = \lambda(1+\lambda).$$
We thus have
$$\frac{AP}{PD} = \frac{BQ}{QE} = \frac{CR}{RF}.$$
This equality implies
$$\frac{AD}{PD} = \frac{BE}{QE} = \frac{CF}{RF}.$$
Using $AD = BE = CF$, we get $PD = QE = RF$. But then $AP = BQ = CR$ and hence $AR = BP = CQ$. Consider the triangles $ABP$ and $BCQ$. Observe that $\angle BPA = 120^\circ = \angle CQB$, $BP = CQ$ and $AP = BQ$. Thus $ABP$ and $BCQ$ are congruent triangles. This gives $AB = BC$. Similarly, we obtain $BC = CA$.

Let us write $PQ = QR = RP = x$, $BE = CF = AD = y$, $AR = u$, $BP = v$ and $CQ = w$. Using the collinearity of the points $A$, $F$, $B$ with respect to the triangle $PQR$, Menelaus' theorem gives
$$\frac{RA}{AP} \cdot \frac{PB}{BQ} \cdot \frac{QF}{FR} = 1.$$
(we are using only the lengths). This may be written in the form
$$\frac{u}{u+x} \cdot \frac{v}{v+x} \cdot \frac{y-w}{y-(w+x)} = 1.$$
Simplification gives
$$u+v = \frac{(x+u)(x+v)(x+w) - uvw}{yx} - x.$$
Note that the expression on the right side is symmetric in $u, v, w$. It follows that
$$u + v = v + w = w + u.$$
Thus $u = v = w$ and hence $AP = BQ = CR$. As in the earlier solution, it follows that the triangles $ABP, BCQ, CAR$ are congruent, and hence $AB = BC = CA$.