from the hint given by @Robertbryant, i can answer the question without computation. i will fix the coordinate system with the center at the origin with radius one and the point $d = 1$. the other four points have $a, b, c$ and $d.$

let $c$ and $a$ approach $b.$ considering the triangle $abe$ the center is $\frac{2}{3}a + \frac{1}{3}e$ through this point a line orthogonal to $cd = bd$ is drawn. for the triangle $abc$ the the center is $b$ and line orthogonal to $ed$ is drawn. if you look at a triangle similar to $bed$ but the $1/3$ of the size, the common point of intersection is the orthocenter of this sammler triangle.

from the hint given by @Robertbryant, i can answer the question without computation. i will fix the coordinate system with the center at the origin with radius one and the point $d = 1$. the other four points have $a, b, c$ and $d.$

let $c$ and $a$ approach $b.$ considering the triangle $abe$ the center is $\frac{2}{3}a + \frac{1}{3}e$ through this point a line orthogonal to $cd = bd$ is drawn. for the triangle $abc$ the the center is $b$ and line orthogonal to $ed$ is drawn. if you look at a triangle similar to $bed$ but the $1/3$ of the size anchored at $b$ and a side parallel to $ed.$ then, the common point of intersection is the orthocenter of this smaller triangle.