So △ADE is a regular triangle

So ∠ AED = ∠ DAE = ∠ AED = 60.

= = & gt∠ADB =∠AEC = 180-60 = 120

Because AD=BD

AE=CE

So △ABD and △ACE are isosceles triangles.

It's so easy to get bad

=∠CAE =( 180- 120)/2 = 30

So ∠BAC=60+30+30= 120.

The relationship between straight lines on a plane is only parallel or intersecting.

Assuming that EF and CD do not intersect, EF is parallel to CD, and from the transitivity of parallel lines, EF is parallel to CD and CD is parallel to AB.

Then EF is parallel to AB, which contradicts the intersection of EF and AB.

So EF and CD intersect.

3.∠A = 45°

Suppose ∠DBE=α

Because DE=EB

= = & gt∠BED=∠DBE=α

So ∠AED=∠BED+∠DBE=2α.

Because AD=DE

= = & gt∠A=∠DEA=2α

So ∠CDE=∠A+∠DEA=4α.

So ∠BDC=∠CDE-∠BDE=4α-α=3α.

Because BD=BC

= = & gt∠C=∠BDC=3α

Because AB=AB

= = & gt∠ABC=∠C=3α

In the triangle ∠A +∠C+∠ABC= 180.

= = & gt3α+3α +2α= 180

So α=22.5

So ∠A=2α=45.

4. Because AD = BC

,

AB=DC

So the quadrilateral ABCD is a parallelogram.

So in ∨ BC,

= = & gtDE∨BF

because

DE=BF

So the quadrangle BEDF is a parallelogram.

= = & gtBE=DF。

What you don't understand is

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