Let the equilateral triangle be $XYZ$. We show that there exists a point on some side that has the same colour as its projection to another side. For that, take points $P$, $Q$ and $R$ on sides $XY$, $YZ$ and $ZX$, respectively, in such a way that $XP : XY = YQ : YZ = ZR : ZX = 1 : 3$ (Fig. 17).

Then $PQ \perp YZ$ because, denoting the midpoint of $YZ$ by $T$, $TQ : TY = (\frac{1}{2} - \frac{1}{3}) : \frac{1}{2} = 1 : 3 = XP : XY$, implying $PQ \parallel XT$. Hence $Q$ is the projection of $P$ to $YZ$. Analogously, $R$ is the projection of $Q$ to $ZX$ and $P$ is the projection of $R$ to $XY$. As at least two points among $P$, $Q$ and $R$ must have the same colour, a point and its projection have the same colour.

W.l.o.g., let $P$ and its projection $Q$ both be red. Let $M$ be the projection of $Q$ to $XY$ and $N$ the projection of $M$ to $YZ$. If $M$ is red, then $PQM$ is a right triangle with all vertices red. If $N$ is red, then $PQN$ is a right triangle with all vertices red. If $Y$ is red, then $PQY$ is a right triangle with all vertices red. Otherwise, $MNY$ is a right triangle with all vertices blue.

Fig. 17

Consider a regular hexagon whose vertices lie on the sides of the triangle (Fig. 18). Suppose that two opposite vertices of the hexagon are of the same colour. If there is one more vertex of the same colour among the other four vertices, there is a right triangle with all vertices being of that colour. Otherwise, any three vertices among the four remaining ones form a right triangle with all vertices being of the other colour.

If any two opposite vertices are of different colours, then there exist two neighbouring vertices of different colours. The corresponding opposite vertices are of different colours, too. One pair of these differently coloured vertices lies on a side of the initial triangle; let these be $A$ and $B$ (red and blue, resp.) and their opposite vertices $D$ and $E$ (blue and red, resp.). Angles $ABD$ and $BAE$ are right. Let $S$ be any point on $AB$ that is not a vertex of the hexagon. If $S$ is red, then $SAE$ is a right triangle with all vertices red; if $S$ is blue, then $SBD$ is a right triangle with all vertices blue.

Fig. 18