As shown in the figure, take the midpoint M of BE, connect FM and extend the intersection BC to N, and connect MG.

Then FM =1/2ab =1/2ed = mg.

∴∠MGF=∠MFH

∫FM//AB

∴∠FNC=∠ABC=90

∴∠MFH+∠BHF=90

BH=BG again.

∠BGH=∠BHF

∴∠MGF+∠BGH=90

∠ bgm =∠ d = 90。

∴∠FGH=∠BGH+∠BGM+∠MGF= 180

That is, f, g and h are collinear.