Solution: As shown in the figure, connect BF, AC, CH,

AC = BF，AB=CB，BC=CF，

∴△ABC≌△BCF，

So α = β,

∫AB∨CD，BC∨DG，

∴∠A=∠DCA,∠BCA=∠DGC，

∴△ABC∽△CDG，

α=∠CDG