In oblique hexagon $ABCDEF$ inner angels at vertexes $B, C, E, F$ are equal. Furthermore the equation $AB+DE=AF+CD$ is fulfilled. Prove that line $AD$ and bisectors of the sides $BC$ and $EF$ have commont point (they all cross at one point).
I already know the solution but I also want to know whether my idea is correct:
Let $G$ be the common point of lines $AB$ and $CD$ and $G$ is the common point of lines $AF$ and $DE$. Then triangels $EFH$ and $BCG$ are isosceles. Hence bisector of side $EF$ is once bisector of angle $\angle EHF$ and analogously bisector of side $BC$ is once bisector of angle $\angle BGC$. Because $AB+DE=AF+CD$ and triangels $EFH$ and $BCG$ are isosceles, the equation $AG+DH=AH+DG$ is fullfilled.
And here comes my uncertainty:
Let $K$ be the common point of the bisector of angle $\angle EHF$ and line $AD$. Now my idea is to use the angle bisector theorem and prove that $\frac{BG}{GC}=\frac{BK}{KC}$. It would mean that $K$ lies on the bisector of angle $\angle BGC$ what would finish the proof.
I don't know how to prove that last part.
Do you mean $$\frac{AG}{DG} = \frac{AH}{DH} \, ?$$ Clearly $$\frac{BG}{GC} = 1$$ so what? Do you want to prove that $BK = KC$? Well, maybe the bisector theorem is a bit of an overkill. In order to use it you basically will have to prove that $AG = AH$ and $DG = DH$ which already will imply that $AD$ is in fact the angle bisector of both angles $\angle \, A$ and $\angle D$ and since the quad $AGDH$ is superscribed around a circle, the four angle bisectors of the vertices of $AGDH$, which in this case are just three because $AD$ is common angle bisector for both $A$ and $D$, meet at the center of the incircle.
So basically, you anyway will have to prove $AG = AH$ and $DG = DH$.