Point C is an arbitrary point on the line AB, with AC and BC as sides respectively as regular triangles. Note that the area of △ BCE is S1, and the area of △ ACD is S2 If C is the golden section of AB and AC > BC, S1: S2 is obtained

The area formula of triangle is s = 1 / 2 * AB sinc. Since both triangles are regular triangles, sinc is sin60 ° and AB in each triangle is the same. The difference is that AC > BC. Assuming that the length of AC is m and the length of BC is n, then S1 = 1 / 2 * n * n * sin60 ° and S2 = 1 / 2 * m * m * sin60 ° and M / (M + n) = (√ 5-1) / 2, we can get m / N = (√ 5-1) / (3 - √ 5), so S1: S2 = 11-4 √ 5

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1. As shown in the figure, ab ∥ CD, ∠ ACB = 90 °, e is the midpoint of AB, CE = CD, de and AC intersect at F.; De is perpendicular to AC, and angle ACD = angle As shown in the figure, it is known that AB / / CD, ∠ ACB = 90 °, e is the midpoint of AB, CE = CD, de and AC intersect at F, and the verification is as follows:
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4. What is a ton represented by a symbol
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11. As shown in the figure, the line AB = 4, C is a moving point on the line AB, with AC and BC as sides, the equilateral triangle ACD and equilateral triangle BCE are made, Finding the minimum value of de
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13. Given that the point C is any point on the line AB (C does not coincide with a or b), AC and BC are taken as sides respectively, and the equilateral triangle ACD and equilateral △ BCE, AE and C are made on the same side of ab D intersects with m, BD intersects with CE with n The results show that: (1) ace ≌ DCB; (2) ACM ≌ DCN; (3) Mn ∥ ab
14. As shown in the figure, BD is divided into two parts, be is divided into ABC 2:5, DBE = 21 ° and the degree of ABC is calculated
15. As shown in the figure, BD is divided into two parts, be is divided into ABC 2:5, DBE = 21 ° and the degree of ABC is calculated
16. As shown in the figure, BD is divided into two parts, be is divided into ABC 2:5, DBE = 21 ° and the degree of ABC is calculated
17. As shown in the figure, make equilateral triangles △ abd, △ EBC, △ fac on the same side of BC with the three sides of △ ABC respectively, and prove that the quadrilateral ADEF is a parallelogram
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