(1). Draw an arc with A as the center and AB as the radius, intersect DC at E, and connect BE, then BE is the bisector of ∠AEC ∠.

Proof: According to the drawing, the triangle ABE is an isosceles triangle.

∴? ∠ Abe =∠AEB

And then what? Annie =∠BEC

∴BE is the bisector of ∞∠AEC.

(2).？ ①? In the right triangle ADE, AD=√3AE=2.

Beg? DE = 1 = EC∠DAE = 30 & amp； ordm

A triangle is an equilateral triangle.

In the right triangle ABP, AB=2BP=? √3

Can be obtained? AP=4√3/3？

Then < PAB = 30 &；; ordm？ . ? ∠APB = 60 & amp； ordm。

∠ Angular bisector of ∠BAE and AP and perpendicular bisector of BE.

In the right triangle PEC

By CE= 1CP=？ √3?

Can you get PE= √3=PB

The triangle PBE is an isosceles triangle.

∠PBE =∠PEB = 30 & amp； ordm∠BPF =∠PBE+∠PEB = 60 & amp； ordm

In the triangle PAF, ∠ PAF = 30&; ordm∠APF =∠APB+∠BPF = 120 & amp； ordm

∴∠afp=30&； ordm=∠PAF？

∴? The triangle PAF is an isosceles triangle.

∵? PB⊥AF

∴? Point b bisects line segment AF.

②.？ By 1? We can see this. Triangle PBF≌ triangle PBA≌ triangle pea.

Among them, triangle PBA is mirror congruence with triangle PEA and triangle PBF.

Are triangular PBF and triangular pea identical? .

So what? Triangle PAE can be obtained by rotating triangle PFB clockwise around point P. 。

The degree of rotation is the degree of ∠APF? 120。 ordm。