As shown in the figure, point D is a point on the AB extension line of the equilateral triangle ABC. Take CD as the side length, make the equilateral triangle CDE, and calculate the degree of the angle EBD

As shown in the figure, point D is a point on the AB extension line of the equilateral triangle ABC. Take CD as the side length, make the equilateral triangle CDE, and calculate the degree of the angle EBD

∵ △ ABC and △ CDE are equilateral triangles, ∵ ABC = ∵ CED = ∵ ECD = 60 ° and ∵ B, C, e and D are in common circle,
∴∠EBD＝∠ECD＝60°.

As shown in the figure, it is known that BD and be are the bisector and height of triangle ABC respectively, ∠ ABC = ∠ ACB = 2 ∠ a, and the degree of ∠ EBD is calculated
correct

Angle ABC + angle ACB + angle a = 5, angle a = 180 degree
The results show that angle a = 36 °, angle B = 72 ° and angle c = 72 °
So the angle abd = 1 / 2, the angle B = 36 ° in the triangle Abe, the angle Abe = 90 ° - 36 ° = 54 ° in the triangle Abe
Finally, the angle EBD is 54 ° - 36 ° = 18 °

As shown in the figure, in the triangle ABC, the angle BAC is equal to 90 degrees, and the angle ACB = 2

∵ AE bisection ∠ BAC, AF bisection ∠ CAD
∴∠EAF=∠EAC+∠CAF=1/2(∠BAC+∠CAD)=90°
The Δ EAF is a right triangle
∵∠ACB-∠B=90°
∴∠BAC=180°-∠ACB-∠B=180°-(90°+∠B)-∠B=90°-2∠B
∴∠BAE=1/2∠BAC=45°-∠B
∴∠AEC=∠BAE+∠B=45°
The EAF is an isosceles right triangle

Know the inscribed circle of ABC, the tangents of each side of triangle and the point def, and the angle FOD is equal to the angle EOD is equal to 135 degrees, try to judge the shape of angle ABC

It's an isosceles right triangle
In quadrilateral adof, angle ADO = angle AFO = 90, angle DOF = 135, so angle a = 45 degree
Similarly, angle B = 45 ° and angle c = 90