∠∠5 =∠ 1+∠2+∠B

? = (∠ 1+∠B)+∠2

= ∠4+∠2

= (∠2+∠3)+∠2

? =∠ 1+∠2+∠3

Namely: ∠ACF=∠BAF

Again: AFC = BFA.

∴△ACF∽△BAF (two equilateral triangles are similar)

Proof: (Figure 2)

∫≈ 1+∠ABC = 90

∠A+∠ABC=90

∴∠ 1=∠A

∵CD⊥AB CF⊥BE

∴ Four-point * * circle of B, C, F and D (four-point * * circle if both ends of the line segment are at the same side and have equal angles)

? (Attachment: BC line segment is stretched into two equal right angles on the same side)

∴∠∠ 1 = ∠ 2 (in the same circle, the circumferential angles of the same chord are equal)

And <1= < a.

∴∠2=∠A

∠ Abe =∠ FBD again

∴△AEB∽△FDB (two equilateral triangles are similar)