As shown in the figure, in the square ABCD, f is a point on CD, AE ⊥ AF, point E is on the extension line of CB, EF intersects AB at point G, proving: de · FC = BG · EC

As shown in the figure, in the square ABCD, f is a point on CD, AE ⊥ AF, point E is on the extension line of CB, EF intersects AB at point G, proving: de · FC = BG · EC

It can be seen from the meaning of the title
DE>EC
FC>BG
So de · FC > BG · EC
Is the title wrong