Iterations of orthic triangles

Nice question! The following is an answer to the first part concerning convergence to a point. (I'm not sure about the behaviour of the angle sequence.)

Theorem. In all non-degenerate cases, the sequence of iterated orthic triangles indeed converges to a point.

(I'll comment on the degenerate case at the end.)

The two key properties of the (non-degenerate) orthic triangle are the following. Consider (non-right) triangle $T = \triangle ABC$ and let $T'$ be its orthic triangle. Then,

1. the circumradius of $T'$ is exactly half that of $T$;
2. the closed discs bounded by the circumcircles of $T$ and $T'$ intersect.

These together suffice to show that any sequence of points in the iterated orthic triangles is Cauchy, hence converges and its easy to then show that the limit point does not depend on the particular sequence of representative points.

Property 1 is asserted all over the place but most of the freely available literature is not clear about whether it holds for obtuse triangles so let me give a proof due to Graeme Wilkin:

Proof of property 1. The claim of property 1 is obvious for equilateral triangles. In all other cases, let's help ourselves to the following facts (see, e.g., Posamentier and Ingmar, The Secrets of Triangles for elementary proofs).

• The well-known nine-point circle theorem: for any triangle, the feet of the altitudes, the midpoints of the sides, and the midpoints of the segments joining each vertex to the orthocenter (the intersection point of the altitudes) all lie on a common circle (the nine-point circle).
• The center of the nine-point circle bisects the segment joining the circumcenter and orthocenter. (This is a non-degenerate segment unless the triangle is equilateral.)

Let $O$, $H$, $N$ be the circumcenter, orthocenter, and center of the nine-point circle of $T = \triangle ABC$ respectively. (If $T$ is obtuse, say $A$ is the obtuse angle.) Let $E_A$ be the midpoint of $HA$. By definition, $|HE_A| = \frac12|HA|$. By the nine-point circle theorem, the circumradius of the orthic triangle $T'$ is $|NE_A|$ and since $N$ bisects $OH$, he have that $|HN| = \frac12|HO|$. Thus, $\frac{|HE_A|}{|HA|} = \frac12 = \frac{|HN|}{|HO|}$ and by the intercept theorem, it follows that $NE_A$ is parallel to $AO$. Then using similar triangles $\triangle HNE_A$ and $\triangle HAO$, we have that $|NE_A| = \frac12|OA|$. But $|OA|$ is the circumradius of $T$! Thus, the circumradius of $T'$ is exactly half that of $T$. QED.

Proof of property 2. Let $D$, $D'$ be the closed discs bounded by the circumcircles of $T$ and $T'$, respectively. Obviously, $T \subset D$ and $T' \subset D'$. But also, at least one vertex, call it $A'$, of $T'$ is contained in $T$ (they all are if $T$ is acute, the foot of the altitude from the obtuse angle is if $T$ is obtuse.) So $A' \in D' \cap D$ and the discs intersect as claimed. QED.

Now we can prove the convergence result.

Proof of theorem. Let $T_0,T_1,T_2,\dots$ be the iterated orthic triangle sequence $T_{n+1} = T_n'$ for some initial $T_0$ and assume none of them are degenerate. (For the following, you may take these triangles to be either the 1-skeletons or the closed filled polygons, it doesn't make a difference.) Take arbitrary $x_n \in T_n$ for each $n$. Let $r_n$ be the circumradius of $T_n$. By property 1, we have that $r_{n+1} = \frac12 r_n$. By property 2, take $w_n$ to be a point belonging the circumdiscs of both $T_n$ and $T_{n+1} = T_n'$. Then $$ \lVert x_n - x_{n+1} \rVert \le \lVert x_n - w_n\rVert + \lVert w_n - x_{n+1}\rVert \le 2r_n + 2r_{n+1} = 3r_n = \frac{3r_0}{2^n}. $$ Therefore, the sequence $(x_n)$ is Cauchy, hence converges, say to $x$.

Lastly, let's see that this limit point $x$ does not depend on the particular choices of the $x_n$ in each $T_n$. Consider some other sequence $y_n \in T_n$. By the same reasoning, $y_n \to y$ for some point $y$. But for the same reason again, the interleaved sequence $x_0,y_1,x_2,y_3,x_4,y_5,\dots$ also converges, hence $x = y$. QED.

Remark. If you prefer, one can also dispense with these sequences of points and argue instead that the sequence of triangles itself is Cauchy in a certain complete metric space, but the reasoning is essentially the same when one unpacks the definitions. Namely, consider the sequence of triangles (again either the 1-skeletons or the closed filled polygons) as elements of the space $F(\mathbb R^2)$ of all non-empty compact subsets of $\mathbb R^2$ equipped with the Hausdorff metric $d_{\mathrm H}$. Since $\mathbb R^2$ is complete, so is $(F(\mathbb R^2),d_{\mathrm H})$. Reasoning as above, one finds $(T_n)$ is Cauchy w.r.t. $d_{\mathrm H}$, hence converges to some non-empty compact set $X$. But also, using the circumradii, one argues that the diameter of $X$ is zero, hence $X$ is a singleton.

The degenerate case. It seems to me that how you handle the degenerate case is a matter of taste. Obviously, if you simply terminate the iterative process whenever you hit a degenerate triangle, then in these cases, you won't have convergence to a point. But I can imagine reasonable conventions for the orthic triangle of a degenerate triangle which make both the theorem and the above proof valid in all cases. If $\triangle ABC$ is a right triangle with right angle $A$, then the degenerate orthic triangle is $\triangle AH_AA$ where $H_A$ is the foot of the altitude from $A$, so the degenerate case concerns triangles of the from $\triangle PQP$. Of course, the "side" $PP$ does not determine a line, so there is no altitude from $Q$ and (perhaps more seriously) there isn't a unique limit of orthic triangles for $\triangle PQR$ as $R \to P$. However, there is a unique limiting nine-point circle (passing through the midpoints of $PQ,QR,RP$) converging to the circle $S$ whose center lies in the segment $PQ$ and passes through $P$ and the midpoint $M$ of $PQ$. Moreover, if we regard this circle $S$ as the nine-point circle of $T = \triangle PQP$ and insist that the orthic triangle $T'$ is inscribed in $S$, then we preserve the circumradius halving feature. These conditions still don't uniquely determine a triangle $T'$ but we might, by fiat, insist (like the acute case) that $T'$ is also inscribed in $T$. This leaves only the choices $\triangle PPP$ or $\triangle PMP$. I personally like the latter because it seems wrong to say that $\triangle PPP$ is inscribed in $S$. Thus, I propose the convention: the orthic triangle of $\triangle PQP$ is the triangle $\triangle PMP$ where $M$ is the midpoint of $PQ$ and the circumcircle of $\triangle PQP$ is the circle with center $M$ and radius $MP$. For completeness, we might also say that the super degenerate $\triangle PPP$ is its own orthic and has circumcircle with center $P$ and radius zero. Under these conventions, the above proof works in all cases (including $T_0 = \triangle PPP$).