![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
Taking one side CD of square ABCD as equilateral triangle CDE, then ∠ AEB=-----
Convenient point, let the side length be 2, make the center line of the CD side of the triangle, and the midpoint F
The midline extends to edge g of AB, where G is the midpoint of ab
EF=√3,BE=2-√3,AE²=BE²=1+（2-√3）²
Cosine theorem
cos∠AEB=（1+（2-√3）²+1+（2-√3）²-2²）/2（1+（2-√3）²）（1+（2-√3）²）
It is reduced to cos ∠ AEB = - √ 3 / 2
That is, AEB is 120 degrees
RELATED INFORMATIONS
- 1. Given that the quadrilateral ABCD is a square and E is a point in the square, △ ade is an equilateral triangle, find the degree of ∠ EBC
- 2. As shown in the figure, in the isosceles trapezoid ABCD, e is the midpoint of the bottom BC, connecting AE and de
- 3. As shown in the figure, in the isosceles trapezoid ABCD, e is the midpoint of the bottom BC, connecting AE and de
- 4. As shown in the figure, in isosceles trapezoid ABCD, ad ‖ BC. AB = DC, e is the midpoint of BC, connecting AE and De, proving: AE = De
- 5. As shown in the figure, in the isosceles trapezoid ABCD, ad ∥ BC, point E is the midpoint of BC side
- 6. As shown in the figure, in the right angle trapezoid ABCD, ad ‖ BC, ∠ C = 90 °, ad = 5, BC = 9, take a as the center, rotate waist AB clockwise 90 ° to AE, connect De, then the area of △ ade is equal to () A. 10B. 11C. 12D. 13
- 7. As shown in the figure, in the square ABCD, the diagonal AC and BD intersect at point E, AF bisects ∠ BAC, intersects BD at point F, and proves: EF + 12ac = ab
- 8. As shown in the figure, in rectangular ABCD, the diagonal lines AC and BD intersect at point O, AE bisects ∠ bad, BC intersects at point E. if ∠ CAE = 15 °, calculate the degree of ∠ BOE
- 9. As shown in the figure, in rectangular ABCD, the bisector of ∠ bad intersects BC at point E, and point O is the intersection of diagonals, and ∠ CAE = 15 °, then ∠ BOE=______ Degree
- 10. It is known that: as shown in the figure, the diagonal of rectangle ABCD intersects at O, AE bisects ∠ bad intersects BC at e, ∠ CAE = 15 °, then ∠ BOE=______ °．
- 11. As shown in the figure, if the quadrilateral ABCD is a square and △ CDE is an equilateral triangle, then ∠ AED=______ ，∠AEB=______ ．
- 12. Quadrilateral ABCD is a square, triangle CDE is an equilateral triangle. Find the degree of angle AEB I can't send the picture, but describe it. A square (adce in clockwise) is connected with an equilateral triangle Dec on the left
- 13. As shown in the figure, in the parallelogram ABCD, AE is perpendicular to BC and E, AF is perpendicular to CD and F, the angle BAE = 30 degrees, be = 2, CF = 1. 1) calculate the area of triangle ECD; 2) if ed and AF intersect g, calculate the length of eg
- 14. As shown in the figure above, e is a point on the extension line of the edge ab of the parallelogram ABCD, and de intersects BC with F. prove that the area of triangle ABF = the area of triangle efc The picture can't be transmitted
- 15. In a parallelogram ABCD, e is a point on the extension line of edge AB, and de intersects BC with F. it is proved that s △ ABF = s △ efc
- 16. E is a point on the extension line of the edge ab of the parallelogram ABCD, and de intersects BC with F. it is proved that s △ ABF = s △ efc
- 17. E is a point on the extension line of the side ab of the parallelogram ABCD. De intersects BC at point F. it is proved that the area of triangle ABF and triangle EFC are equal
- 18. Point E is a point on the extension line of the edge ab of the parallelogram ABCD, and de intersects BC with F. proof: s △ ABF = s △ efc
- 19. In the quadrilateral ABCD, the diagonal AC bisector angle bad, AB > ad proves that ab-ad > cb-cd
- 20. As shown in the figure, in diamond ABCD, e and F are the points on CB and CD respectively, and be = DF. If AC is connected, can you determine the relationship between AC and ef