Inspired by this question I began thinking about the following:
Given a quadrilateral ABCD where $\triangle ABC$ is a fixed isosceles triangle and D is free to move, where can D be placed such that $\triangle AED$ is similar to $\triangle ADC$? The figure above gives an example of such a position.
Using trial and error (iteration) I was able to map out such positions:
As you can see, the positions turn out to give a rather nice (half) heart shaped curve.
My question is: Can anyone work out what the equation of this curve is?
Using your picture from above, set $2k = \overline{AC}$, $h = \textrm{dist}(B, AC)$, and $\theta = \angle AED$.
For fixed $k, h, \theta$, there clearly is a unique point $D$ such that the two triangles $\triangle AED$ and $\triangle ADC$ are similar.
Set $a := \overline{AE} = k - h\cot(\theta), s := \overline{DE}$. By the cosine law, $\overline{AD}^2 = a^2 + s^2 - 2as\cos(\theta)$. Also, similarity says that $$\frac{\overline{AE}}{\overline{AD}} = \frac{\overline{AD}}{\overline{AC}},$$ or $a^2 + s^2 - 2 as \cos(\theta) = 2ak$, so that
\begin{align*} s &= a \cos(\theta) + \sqrt{ a^2\cos^2(\theta) + a(2k-a)}\\ &= (k - h \cot(\theta))\cos(\theta) + \sqrt{(k - h \cot(\theta))^2\cos^2(\theta) + k^2 - h^2 \cot^2(\theta)} \end{align*}
Unfortunately, there seems to be no simpler expression for $s$ (that I could find, at least). In any case, we now have a closed form expression for $D$ in polar coordinates by centering at $B$ and noting that $$r := \overline{BD} = \frac{h}{\sin(\theta)} + s.$$
Hope this helps (but I don't think this would look like a cardioid in general).