Let ∠DBC=X degrees

∫DC = CB

∴∠CDB=∠CBD=X

DC AB

∴∠CDB=∠DBA=X

∴∠CBA=2X

∵ trapezoid ABCD is an isosceles trapezoid.

∴∠DAB=∠CBA=2X

DB = AB

∴∠DAB=∠ADB=2X

Use triangles and 180 degrees in triangle ADB.

X+2X+2X= 180

The solution is X=36.

2X=72 degrees

So the angle a is 72 degrees.

2.

In order to ensure that the distance to three o'clock is equal, we must first make the distance to two points B and C equal.

The point with equal distance to two points is on the perpendicular line of the line segment formed by these two points.

So perpendicular bisector, who is line BC,

Similarly, the points with equal distances to A and C are also on the middle vertical line of line segment AC.

Two perpendicular bisector intersect at one point.

The intersection is the location of the transfer station.

The ruler draws the middle vertical line of the line segment.

3.

Solution: DEC is an equilateral triangle and ABCD is a square.

∴EC=DC=BC ∠BCE=90+60= 150

∴∠EBC=∠CEB= 15

∴∠ABE=90- 15=75

In triangle ADE and triangle BCE

AD=BC DE=CE ∠ADE=∠BCE

∴△ADE≌△BCE

∴AE=BE

Triangle ABE is an isosceles triangle with a base angle of 72 degrees.

So the vertex angle, that is,

∠AEB=36 degrees

4.

Your picture is wrong first, and then draw the following picture.

Solution: △ADC' is the AD folding diagram of △ △ACD delay line.

∴△ADC≌△ADC'

∴∠ADC=∠ADC'=60 CD=DC '

And ∵D is the midpoint of BC.

∴BD=DC'

∠BDC'=60 degrees

△BDC' is an equilateral triangle.

BC'=BD=DC=2

5.

Are all right.

Options a and b directly use the properties of equilateral triangles.

C, equilateral triangle is a special isosceles triangle, so it also has the nature of three lines in one.

So AE is the symmetry axis.

D, the midline and high line of either side of an equilateral triangle coincide with the bisector of an angle.

So AE is the bisector of ∠BAC.