I was solving this problem :
Reconstruct the triangle from the points at which the extended
bisector, median and altitude drawn from a common vertex intersect the circumscribed circle.
I found the solution to this problem but I wonder since we can always construct the triangle back from these three points what are the relations between the different triangles that can be constructed by swapping the points. i.e given the three points $E,F,G$ In the triangle $ABC$ let the $E$ be the point that intersects the extended bisector and the circumscribed circle, let $F$ be the point that intersects the extended altitude and the circumscribed circle, let $G$ be the point that intersects the extended median and the circumscribed circle,In the triangle $XYZ$ let the $E$ be the point that intersects the extended bisector and the circumscribed circle, let $G$ be the point that intersects the extended altitude and the circumscribed circle, let $F$ be the point that intersects the extended median and the circumscribed circle.
For some reason the point that intersects the extended bisector and the circumscribed circle is always between the point that intersects the extended altitude and the circumscribed circle and the point that intersects the extended median and the circumscribed circle (I couldn't rigorously prove why ), so there are only $2!$ triangles which could swap the points $F,G$.
Before I drew these triangle I thought that they will be similar triangles but I was wrong, But I still think that these two triangle must have some interesting relation between them as for any triangle there is a unique triangle that can be obtained with this method.
I will define two triangle that are constructed this way by permutated triangles.
Another question If given the two triangles $ABC$ and $XYZ$ how to determine if they are permutated triangles?
Best Answer
Given $\triangle ABC$ in its circumcircle, we can construct companion $\triangle XYZ$ by following $A$ through midpoint $M$ of $\overline{BC}$ to point $G$ on the circle; then reflecting $\triangle GBC$ in the diameter parallel to $BC$:
That the angle bisectors at $A$ and $X$ meet at $E$, the common midpoint of arcs $\stackrel{\frown}{BGC}$ and $\stackrel{\frown}{YGZ}$, is clear (though not shown above). That the altitude-line from $A$ meets the median-line from $X$ at common point $F$ is apparent from the reflective symmetry of the elements relative to the specified diameter.
Notes.
Because $\triangle ABC$ and $\triangle XYZ$ combine to form a cyclic quadrilateral, we know that $\angle A$ and $\angle X$ are supplementary; as are $\angle B+\angle Y$ and $\angle C+\angle Z$.
Moreover, sides $a:=|BC|=|YZ|$, $b:=|CA|$, $c:=|AB|$, $y:=|ZX|=|GC|$, $z:=|XY|=|GB|$, and medians $m:=|AM|$ and $w:=|XW|=|GM|$ (whoops! I forgot to label midpoint $W$ in my figure), are related by Ptolemy's Theorem: $$bz+cy = a(m+w) \tag1$$
Calculating the power of point $M$ with respect to the circle in two ways, we conclude: $$a^2 = 4mw \qquad\to\qquad \frac{2m}{a}=\frac{a}{2w} \tag2$$
Further, because $AG$ cuts $\square ACGB$ into equal-area triangles, we have $$b y \sin\angle ACG = c z\sin\angle ABG \qquad\to\qquad \frac{b}{z}=\frac{c}{y} \tag3$$ So, while the triangles aren't (usually) similar, two of the three pairs of sides are proportional. If you reflect $Z$ in $X$ to get $Z'$, then $\triangle ABC$ is similar to $\triangle XZ'Y$. (I wonder if there's an "obvious" reason for this.)
I'll leave it as an exercise to the reader to show that the ratios in $(2)$ and $(3)$ are all equal.