Solution:
Let $AD = h$, $PD = y$ and $BD = DC = a$. We

observe that $BP^2 = a^2 + y^2$. Moreover,
$PE = PA \sin \angle DAC = (h - y) \frac{DC}{AC} = \frac{a(h - y)}{b}$,
where $b = AC = AB$. Using $AP / PD = (h - y) / y$, we obtain $y = h / (1 + \lambda)$. Thus
$$\lambda^2 = \frac{BP^2}{PE^2} = \frac{(a^2 + y^2) b^2}{(h - y)^2 a^2}$$
But $(h - y) = \lambda y = \lambda h / (1 + \lambda)$ and $b^2 = a^2 + h^2$. Thus we obtain
$$\lambda^4 = \frac{(a^2(1 + \lambda)^2 + h^2)(a^2 + h^2)}{a^2 h^2}$$
Using $m = a / h$ and $z = m^2(1 + \lambda)$, this simplifies to
$$z^2 - z(\lambda^3 - \lambda^2 - 2) + 1 = 0$$
Dividing by $z$, this gives
$$z + \frac{1}{z} = \lambda^3 - \lambda^2 - 2$$
However $z + (1 / z) \geq 2$ for any positive real number $z$. Thus $\lambda^3 - \lambda^2 - 4 \geq 0$. This may be written in the form $(\lambda - 2)(\lambda^2 + \lambda + 2) \geq 0$. But $\lambda^2 + \lambda + 2 > 0$. (For example, one may check that its discriminant is negative.) Hence $\lambda \geq 2$. If $\lambda = 2$, then $z + (1 / z) = 2$ and hence $z = 1$. This gives $m^2 = 1 / 3$ or $\tan (A / 2) = m = 1 / \sqrt{3}$. Thus $A = 60^\circ$ and hence $ABC$ is equilateral.

Conversely, if triangle $ABC$ is equilateral, then $m = \tan (A / 2) = 1 / \sqrt{3}$ and hence $z = (1 + \lambda) / 3$. Substituting this in the equation satisfied by $z$, we obtain
$$(1 + \lambda)^2 - 3(1 + \lambda)(\lambda^3 - \lambda^2 - 2) + 9 = 0$$
This may be written in the form $(\lambda - 2)(3\lambda^3 + 6\lambda^2 + 8\lambda + 8) = 0$. Here the second factor is positive because $\lambda > 0$. We conclude that $\lambda = 2$.