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。