Bumppppp

What does Equilateral Trapezoid mean? Wont it become a rhomus?

Yup, the trapezium is actually a rhombus (unless the OP meant an isosceles trapezium which is different).

the same formulation is also here in English, I shall find a non - English source

From 1999-2000: Problems and Solutions from Around the World by Titu Andreescu page 112, the wording is the following: Quote: Let ABCD be an isosceles trapezoid with bases AB and CD. The incircle of the triangle BCD touches side CD at a point E. Let F be the point on the internal bisector of \DAC such that EF $\perp$ CD. The circumcircle of triangle ACF intersects the line CD at two points C and G. Show that triangle AFG is isosceles. , so I shall correct it, replacing the ''equilateral trapezoid with sides'' with ''isosceles trapezoid with bases''

parmenides51 wrote: Let $ABCD$ be an isosceles trapezoid with bases $AB$ and $CD$. The incircle of the triangle $BCD$ touches $CD$ at $E$. Point $F$ is chosen on the bisector of the angle $DAC$ such that the lines $EF$ and $CD$ are perpendicular. The circumcircle of the triangle $ACF$ intersects the line $CD$ again at $G$. Prove that the triangle $AFG$ is isosceles. Let $I_2$ be the incenter of $ACD$ and $M$ the midpoint of arc $\widehat{CD}$ not containing $A$.Then by fact 5 it follows $\odot{CDII_2F}$ all lie on circle with center $M$ and $F$ is the antipode of $I_2$ in this circle.Now since $\angle AGF=180^{\circ}-\angle ACF=\angle AI_2D-\angle ACD=90^{\circ}-\angle \dfrac{C}{2}=\angle DI_2F=\angle FAG$ so $AF=AG$ and we are done.