It is proved that FH at point F is parallel to the parallel line where AB intersects BC at point H and AF at points D and H. ∫AF∨DH.

, AB parallel FH

∴ Quadrilateral ADHF is a parallelogram

Namely AF=DH

AF = BD = BH

∴△DBH is a regular triangle

That is, ∠ B = ∠ A = 60, ∴△ABC is a regular triangle (an isosceles triangle with an angle of 60 is an equilateral triangle).