It is known that C is a point on the line AB, and △ ACD and △ BCE are equilateral triangles

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• 1. As shown in the figure, C is a point on the line AB, △ ACD and △ BCE are equilateral triangles. AE intersects CD at point m, BD intersects CE at point n, AE intersects AE at point O. The results are as follows: (1) AOB = 120 °; (2) cm = cn; (3) Mn ‖ ab
• 2. In the quadrilateral ABCD, BC = 2, DC = 4, and ∠ A: ∠ ABC: ∠ C: ∠ ADC = 3:7:4:10, find the length of ab
• 3. As shown in the figure, in △ ABC, a = 70 ° and B = 50 °, CD bisects ACB, and calculates the degree of ACD
• 4. In the triangle ABC, the bisector of the angle ABC intersects the bisector of the angle ACD at e, and the angle a is equal to 78 degrees
• 5. As shown in the figure, ∠ ACD = 60 ° and ∠ BCD = 40 °, CE and CF are the bisectors of △ ACD and △ BCD respectively, and the degree of ∠ ECF can be calculated
• 6. It is known that as shown in the figure, ∠ ACD is an external angle of △ ABC, CE and CF divide ∠ ACB, ∠ ACD, eh ‖ BC equally, and intersect AC and CF at points g and h respectively. Verification: Ge = GH
• 7. CE and CF are the bisector of the inner angle and the bisector of the outer angle of the triangle ABC, respectively
• 8. As shown in the figure, in △ ABC, ∠ a = 70 ° bisector CE ‖ ab of outer angle, calculate the degree of ∠ B and ∠ ACB
• 9. As shown in the figure, in the triangle ABC, ∠ a = 40 °, CE is the bisector of the outer angle of ∠ ABC, and CE ‖ AB, calculate the degree of ∠ B and ∠ ACB
• 10. Let a = (COS α, sin β) and B = (COS β, sin α), then α - β =? 0
• 11. In the quadrilateral ABCD, ab = 2, radical 5, BC = 2, CD = 1, ad = 5, and angle c = 90 degrees, find the area of quadrilateral ABCD
• 12. E is a point in the square ABCD, and the angle ade is equal to the angle DAE is equal to 15 degrees. It is proved that BCE is an equilateral triangle
• 13. E is a point in the square ABCD, and there is ∠ ade = ∠ DAE = 15 ° to prove that △ BCE is an equilateral triangle
• 14. E is a point in the square ABCD, and △ EAB is an equilateral triangle, then ∠ ade=______ ．
• 15. E is a point in the square ABCD, and △ EAB is an equilateral triangle, then ∠ ade=______ ．
• 16. A special parallelogram problem is known as rectangle ABCD, which makes an equilateral triangle ade and an equilateral triangle CDF with AD and CD as one side respectively, connecting be and BF The value of be / BF is equal to_____ .
• 17. Taking the adjacent sides AB and ad of the parallelogram ABCD as one side, we make an equilateral triangle ABF and ade outside the parallelogram, and prove that △ CEF is also an equilateral triangle
• 18. In the parallelogram ABCD, ad and BC are taken as sides respectively, and the equilateral △ ade and equilateral triangle BFC are made outside the parallelogram ABCD
• 19. There are 500 kg apples in the fruit shop, just 1 / 9 more than pears. How many kg pears?
• 20. The total weight of two cases of apples is 20 kg. If 4 / 1 kg is taken out of the first case and put into the second case, the two cases of apples will have the same weight? Please