Why are they congruent?

Question

Let $M$ be the orthocentre of $\triangle ABC$. Furthermore let $X, Y, Z$ be the circumcenters of triangles $BCM, ACM, ABM$.

Prove that triangles $ABC, XYZ$ are congruent. I have proved that $\triangle ABC, \triangle XYZ$ are similar, but I don't know how to prove that they have a same side...

Answer 1

It is ''well known'' that the circle around $AMB$ is congruent to circle around $ABC$.

Since $XM=YM=ZM=R$ the $M$ is circumcenter of $XYZ$ with the same radius as circumcircle $ABC$. So since they are similar they must be also congruent.

Answer 2

$AYMZ$ is a rhombus for which in the standard notation $AM=a|\cot\alpha|$ and $AY=R$.

Thus, $$\frac{1}{2}YZ=\sqrt{R^2-\frac{1}{4}a^2\cot^2\alpha}=R\sqrt{1-\cos^2\alpha}=R\sin\alpha=\frac{a}{2}$$ and we are done!

Answer 3

Let $C(XYZ)$ denote the circumscribed circle to triangle $XYZ$.

A main remark is that circles $C(MAB)$, $C(BMC)$ and $C(MCA)$ have a same common radius, which is the radius of $C(ABC).$

This is an easy consequence of the classical theorem (recalled by @Jack D'Aurizio) that the symmetric of orthocenter $H$ with respect to any side of triangle $ABC$ belongs to $C(ABC)$ ; thus $C(MAB)$ is symmetrical of $C(ABC)$ w.r.t. to side $AB$, thus has the same radius. Similar reasoning for the two other circles $C(BMC)$ and $C(MCA)$.

As these three circles have the same radii, the result is a direct consequence of the so-called Johnson's circles theorem : see property 5 in (https://en.wikipedia.org/wiki/Johnson_circles)

Answer 4

If $H$ is the orthocenter of $ABC$, the vertices of $XYZ$ can be found by intersecting the perpendicular bisectors of $HA,HB,HC$. By applying a dilation with center $H$ and ratio $2$ we have that $XYZ$ is mapped into the anticomplementary triangle of $ABC$.

This problem can also be solved by recalling that the symmetric of $H$ with respect to a side of $ABC$ lies on the circumcircle of $ABC$ (by angle chasing), or by recalling that the nine-point-circle of $ABC$ goes through the midpoints of $HA,HB,HC$. We also have that $AX,BY,CZ$ concur in the center of the nine-point-circle of $ABC$.