## Concurrent Congruent Segments

#### Problem

In triangle $ABC$ equilateral triangles $ABX$, $BCY$, and $ACZ$ are constructed.

Prove that $AY = BZ = CX$ and that all three lines are concurrent at $O$.

#### Solution

We shall begin by showing that the three segments are congruent.

Consider triangle $AXC$ in the following diagram.

By rotating edge $AX$ through 60 degrees about $A$ we produce edge $AB$ and edge $AC$ coincides with edge $AZ$ (also rotated 60 degrees about $A$). As $\angle CAX$ is preserved through this rotation, $\angle CAX = \angle BAZ$. Therefore triangles $AXC$ and $ABZ$ must be congruent and it follows that $CX = BZ$.

By considering triangle $BCX$ and triangle $BAY$ in the same way we can show that $AY = CX$.

Thus we show that $AY = BZ = CX$.

To prove that the segments are concurrent at $O$ we shall consider the circumscribed circles, with $X'$, $Y'$, and $Z'$ representing the circumcentres of each triangle.

Clearly circles $X'$ and $Y'$ intersect at $B$ and $O$, but we must also show that circle $Z'$ passes through this common point.

As $OAXB$ is a cyclic quadrilateral the sum of opposite angles is 180 degrees. As $\angle AXB = 60$ degrees, $\angle AOB = 120$ degrees.

Similarly $\angle BOC = 120$ degrees.

As the angles around $O$ add to 360 degrees, it follows that $\angle AOC = 120$ degrees.

But as $\angle AZC = 60$ degrees, we can use the converse of the cyclic quadrilateral theorem and deduce that because opposite angles, $AOC$ and $AZC$, add to 180 degrees, the points $O$, $A$, $C$, and $Z$ must lie on the same circle. That is, circle $Z'$ passes through $A$ and $C$ and also passes through $O$.

Hence circles $X'$, $Y'$, and $Z'$ are concurrent at $O$.

Now we are able to show that the segments $AY$, $BZ$, and $CX$ are also concurrent at $O$.

In circle $X'$, angles $BAX$ and $BOX$ share the same chord, $BX$, so it follows that $\angle BAX = \angle BOX = 60$ degrees.

In the same way we can show that $\angle BOY = \angle COY = 60$ degrees.

As angles $BOX$, $BOY$, and $COY$ add to 180 degrees, the points $X$, $O$, and $C$ must must lie on the same straight line. Therefore segment $CX$ passes through $O$.

Similarly we can show that segments $AY$ and $BZ$ pass through $O$. **Q. E. D.**

If $P$ represents any point inside triangle $ABC$, the position of the point that minimises $PA + PB + PC$ is called the Fermat/Torricelli point.

- Show that if any interior angle exceeds 120 degrees then the Fermat point is found at that vertex.
- If no interior angle exceeds 120 degrees, prove that the point of intersection of the three circles produced from the constructed equilateral triangles is the Fermat point.
- Prove that this point is unique.