Če nad stranicami trikotnika z zunanje strani narišemo enakostranične trikotnike APB, BQC in CRA, se daljice AQ, BR in CP sekajo v Fermatovi točki Fe trikotnika ABC. Diplomsko delo prinaša nekatere posplošitve tega rezultata. Pri eni nad stranicami narišemo podobne enakostranične trikotnike. Pri drugi nad stranicami na primeren način narišemo podobne (ne nujno enakokrake) trikotnike. Obe omenjeni situaciji sta posebna primera splošnejše situacije, kjer pri vsakem oglišču od obeh stranic, ki se stikata v tem oglišču, navzven odmerimo enake kote. V naslednji posplošitvi iz trikotnika na primeren način ustvarimo šestkotnik in nad stranicami tega narišemo enakostranične trikotnike. Tudi tokrat smo priča dejstvu, da se tri daljice sekajo v skupni točki. Isto se zgodi, če enakostranične trikotnike namesto navznoter narišemo navzven.If we draw equilateral triangles APB, BQC in CRA on the outside of the given triangle ABC, the three segments AQ, BR and CP intersect in the Fermat point Fe of a triangle ABC. In the diploma thesis we present some generalizations of this result. Instead of equilateral triangles we draw similar isosceles triangles. Next we draw (in an appropriate manner) similar scalene triangles. Both mentioned situations are special cases of a more general situation, where at every vertex of a triangle we draw two rays, forming the same angle with the sides of a triangle that meet in that specific vertex. Repeating this in all three vertices with possibly different angles at different vertices, we end up with three triangles APB, BQC in CRA and concurrent lines AQ, BR and CP. Finally we start with a triangle and form a certain hexagon out of it. On the sides of this hexagon we again erect equilateral triangles and end up again with three concurrent lines. The same is true if the equilateral triangles are erected on the inside
