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The "short answer" is as follows. The point $P$ is in general not constructible, for instance, for the triangle with sides $6$, $9$, $13$ constructing $P$ implies we can solve a polynomial equation $\Pi(K)=0$, where $\Pi=a_7K^7+a_6K^6+a_5K^5+a_4K^4+a_3K^3+a_2K^2 + a_1K+a_0$ is an irreducible polynomial of degree seven with rational coefficients, ...
Not an answer. I'm just expanding a comment about @PeterTaylor's observation that the known pseudovertices $X(4)$, $X(74)$, $X(1138)$ lie on the Neuberg cubic ... Bernard Gibert's "Pairs and Triads of points on the Neuberg Cubic connected with Euler Lines and Brocard Axes Isometric Parallel Chords" Proposition 1 characterizes the Neuberg cubic of $\... 0 Here is a proof that is pretty standard for existence and unicity in such geometric setting by convergence, in the figure the initiale cevians intersect at$P$. Arrange the segments to the sides in increasing order, here$PF\le PE\le PG$. It is easy to see that the orange parts are smaller than their corresponding segments$(PE,PF)$and the green ones are ... 0 This is a comment, but too long for that. One can try to apply the$p,q$method, i.e., assume that the three vertices are$(0,0)$,$(1,0)$and$(p,q)$respectively. Then if$P$has barycentric coordinates$(\lambda_1,\lambda_2,\lambda_3)$with respect to the vertices, the equality of the squares of the lengths from$P$to the vertices of the cevian ... 4 This is a report on an unsuccessful computational approach which is rather too long for a comment. I work with complex numbers to represent the points in the obvious way. It suffices to consider$\mu(z) = t(z,0,1)$because this can be extended under the invariants to the full$t(z,z',z'')\$. Since multiplication by a complex number is just rotation and ...