Pole of the golden spiral Starting from a golden triangle (an isosceles triangle with the vertex angle half the basis angle) a well known construction gives a spiral (that approximate a logarithmic spiral), as we can see in this Wikipedia page.
The spiral is constructed starting from a sequence of similar triangle and the figure in this Wikipedia page suggests that the pole of the spiral is the intersection point of the medians of two successive triangles in the sequence, but there is not a proof (or a reference to a proof) of this fact.
Searching in other Internet pages I found some confirms of this fact and also the statement that this pole is  a Brocard point of the triangle, but still without a proof.
I have verified with GeoGebra that those statements seems true, as can be seen in the figure, but I'm not able to find a formal proof.  I can only proof that the two construction of the pole (as medians intersection or as Brocard point) are equivalent, but I'm stuck to prove that the sequence of triangles schrinks exactly to this point .

( In the figure the red lines are the medians and the green circles are the construction of the Brocard point. )
If someone knows the proof or a reference to it I 'll be grateful.
 A: Show that $U$ divides both medians $BQ$ and $DR$ in the same ratio. (There's a straightforward similarity argument.) That'll imply that the "next" median also passes through $U$, etc, etc. Since $U$ is interior to successive triangles in the sequence by this construction, and since the triangles definitely get smaller, the vertices converge to $U$.
Note: There's nothing special about the Golden Triangle insofar as this convergence is concerned. It works for arbitrary isosceles triangles. (But, of course, only the Golden Triangle has the Golden Spiral passing through successive vertices.)
A: Since the pole is internal to a vanishing triangle, then chasing any point internal to a triangle (or on the boundary) will lead towards the pole.
I propose to follow the vertices. Each vertex switches from being a base vertex in a triangle to a top vertex in next iteration triangle.
Considering the triangle $\Delta ABC$ placed on a Cartesian coordinate system, with the following initial conditions: $A(0,0), B(0,b), \hat \theta=180^\circ-\hat B, \text{ the length }AC=s$
Starting at A, generating the next vertex follows this pattern:$$\vec r=b\cdot\sum_{k=0}^\infty {\left(\frac{b}{s}\right)^k\cdot(\cos(k\theta), \sin(k\theta))}$$
From here results the coordinates $x,y$ of the pole:
$$x= b\cdot\sum_{k=0}^\infty {\left(\frac{b}{s}\right)^k\cdot \cos(k\theta)}$$
$$y= b\cdot\sum_{k=0}^\infty {\left(\frac{b}{s}\right)^k\cdot \sin(k\theta)}$$
Final notes
I have used 2D vectors to get to the Cartesian coordinates of the pole. Using complex numbers instead of vectors also works.
There may be some mistakes in vector formula. I appreciate any help at checking and make corrections.
Evaluating the sums is left over.
Hope this helps.
If $z=\frac{b}{s}\cdot e^{i\theta}\Rightarrow x+iy=b\cdot \sum_{k=0}^\infty{z^k}=b\cdot \frac{z}{1-z}\Rightarrow \left \{\begin{align}x&=b\left(1+\frac{\frac{b}{s}\cos\theta-\frac{b^2}{s^2}}{r}\right)\\y&=b\cdot \frac{\frac{b}{s}\sin\theta}{r}\\r&=\left(1-\frac{b}{s}\cos\theta\right)^2-\left(\frac{b}{s}\sin\theta\right)^2\end{align}\right .$
Generating the spiral is connected to generating the next triangle being similar to the current triangle.
The spiral could be generated both towards increasing an vanishing.
The difference between the two sides of spiral generation ties to direct and reverse iterations.
The two iterations-direct and reverse generate the spiral not only in opposite directions, but also in opposite manner between two classes of isosceles triangles: one class defined by the top angle being larger than base angle and the other class by the opposite definition.
In my answer I have presented the solution for the class of isosceles triangles where the top angle is less than base angle.
A: Here is complete answer using complex numbers representation.
We will denote
$$\alpha := \tfrac{\pi}{5} \ \text{and} \ \ \varphi:=\tfrac{1+\sqrt{5}}{2},$$
the golden ratio with its properties :
$$\varphi^2=\varphi+1 \iff \tfrac{1}{\varphi}=\varphi - 1.$$
Also, before proceeding to the solution, let us spend a while on Fig. 1, displaying the initial construction in a kind of pentagonal "alma mater" explaining in a natural way the omnipresence of multiples of $\alpha$ and expressions involving $\varphi$.

Fig. 1 : The association of a regular pentagon and its inscribed pentagram (the star) explain why people have had the idea to consider these triangles with these angles. It helps also to make the connection with the golden ratio $\varphi$. The angular values are denoted by points : one point = $\alpha$, two points = $2 \alpha$, three points = $3 \alpha$.

Fig. 2 : All dimensions have been doubled with respect to Fig. 1. The initial triangle is denoted $A_1B_1C_1$. Its successive images by iterating similitude $s$ given in the text are denoted by $A_kB_kC_k$. The little red star is the "pole".
The idea is to find a similitude transform (composition of a rotation and a homothety = turning and shrinking) that maps $A_kB_kC_k$ onto $A_{k+1}B_{k+1}C_{k+1}.$
The rotation angle (notations of fig. 1) is $$angle(\vec{CA},\vec{AB})=3 \alpha.$$
The homothety (shrinking) ratio is
$$\frac{AB}{CA}=\frac{1}{\varphi}=\varphi - 1.$$
Therefore the complex form $Z=az+b$ of this similitude is :
$$s(z)=\frac{1}{\varphi}e^{3 i \alpha}z+b.$$
It remains to find the value of $b$ with the coordinate system of Fig. 2 where all distances are multiplied by 2 compared to those in figure 1. For example, we have :
$$A=-1, \ B=+1, C=i \tan(2 \alpha), E= i \tan(\alpha), \text{etc.}$$
Let us express that $s(A)=B, i.e., s(-1)=1$.
This constraint allows to obtain $b$, yielding the final form of similitude $s$:
$$s(z)=\frac{1}{\varphi}e^{3 i \alpha}(z+1)+1$$
The looked-for "pole" is the fixed point of this transformation, i.e., the point $p$ such that $s(p)=p$, i.e.,
$$\frac{1}{\varphi}e^{3 i \alpha}(p+1)+1=p \ \iff \ \frac{p-1}{p+1}=u \ \text{with} \ u:=\frac{1}{\varphi}e^{3 i \alpha}\tag{1}$$
If we invert (1), we obtain an explicit expression for the pole:
$$p=\frac{1+u}{1-u}=\frac{1+\frac{1}{\varphi}e^{3 i \alpha}}{1-\frac{1}{\varphi}e^{3 i \alpha}}$$
Now, why is $P$ the (or more exactly one of the two) Brocard point(s) obtained as the "green" circles intersection you give in your figure ?
Because of relationship (1). Indeed taking the argument on both sides of $\frac{p-1}{p+1}=u$ we have
$$3\alpha = \arg(u)=\arg(\tfrac{p-1}{p+1})=\arg(p-1)-\arg(p-(-1))$$
$$=\text{polar angle}(\vec{BP})-\text{polar angle}(\vec{AP})=\text{angle}(\vec{PA},\vec{PB})=\angle(APB),$$ which coincides with the measure $3 \alpha$ of $\angle AEB$, proving that pole $P$ subtends with the same angular value the same arc $AB$ as point $E$.
Otherwise said  $\angle APB = \angle AEB$ proves that point $P$
belongs to the small green circle. Using similitude $s^{-1}$, point $P$ belongs also to the other "green" circle, therefore is situated at their intersection.
