# Why are they congruent?

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...

• Congruent, not "congurrent":) – Dietrich Burde Jan 30 '18 at 16:24
• @Pet123 I have had a look at all your answers : you have validated none of them : this is not the right way to use this site. Once a question has received satisfying answers, you have to validate the answer you consider as the best one for you. Don't await years with the secret wish that somebody will find another (n+1)st interesting answer. A rule of thumb is to validate the best-so-far answer 2 to 4 weeks after you have asked your question. – Jean Marie Feb 21 '18 at 22:39

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. 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 $ABC$.

$AYMZ$ 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!

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)

• Useful information. – Mick Jan 30 '18 at 17:14