are they rays?

Line segments I think

that's not possible

I think it must be possible, problem is from Serbian JBMO TST last year. Maybe wording I gave isn't strict enough. Strict wording is something like this, sorry for my bad English (I used google translate to translate some parts): In acute-angled non-isolaces triangle ABC angle ACB=60°. Let A' and B' be feets of altitudes from vertices A and B respectively and let T be centroid of the triangle. Rays A'T and B'T intersect the circumcircle of triangle ABC in points M and N respectively. Prove that MN=AB. I can post here a sketch later. My idea is to try to prove 2A'T=MT and 2B'T=NT. Because it's well known that because ACB=60° we have A'B'=AB/2.

Try this first $ABC$ is a triangle and $BD$ is an altitude. $T$ is the centroid of $ABC$. Ray $DT$ meets $(ABC)$ at $N$. Prove $AB=NC$.

It's pretty easy and we use this as our lemma. Let $K$ be the midpoint of $AC$, so $BT=2TK$. By our lemma we have $AC=BM=2CK=2A'K$ because $\triangle A'CK$ is equilateral. Then $\frac{BT}{TK}=\frac{BM}{A'K}=2$. By SSA similarity, which works when A is obtuse, $2A'T=MT$.

ythomashu wrote: Try this first $ABC$ is a triangle and $BD$ is an altitude. $T$ is the centroid of $ABC$. Ray $DT$ meets $(ABC)$ at $N$. Prove $AB=NC$. Any hint how to prove this?

We're trying to prove that $ABNC$ is an isosceles trapezoid. Consider the rectangle with vertices at $D,B,$ and $N$ and the lines $DN$ and $BT$.