It is known that in trapezoidal ABCD, AD//BC, AB=CD, E and F are the midpoint of AD and BC respectively, and the extension lines of BA and CD intersect with the extension lines of FE at G point. Description: ∠BGF=∠CGF.

Solution 1: ∫ Quadrilateral ABCD is trapezoidal; AB=CD

So ∠ b = ∠ c.

So BG=CG

So △BGC is an isosceles triangle.

So GF is the bisector of ∠BGC.

So ∠ BGF =∠CGF

Solution 2:

AD//BC，AB=CD？ = & gt？ This is an isosceles trapezoid.

Connect two midpoints e and f vertically.

They all extend the convergence point G into an isosceles triangle, GF is not the bisector of the center line, nor the vertex angle, and the two angles are equal.