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It is known that, as shown in the figure, the quadrilateral ABCD is a parallelogram, and both △ ade and △ BCF are equilateral triangles
It is proved that be and DF are connected by ∵ ▱ ABCD, ∵ ad ∥ BC, ad = BC, (1 point) ∵ ad ∥ BC, (2 points) ∵ equilateral triangle ade, ∵ de = ad, ∵ 3 = 60 degree, (3 points) ∵ equilateral triangle BCF, ∵ BC = BF, ∵ 4 = 60 degree, (4 points) ∵ de = BF, (5 points) ∵ 1 + ∥ 3 = ∥ 2 + ∥ 4, i.e. ∥ BDE = ∥ DBF, ∥ de ∥ BF, (6 points) ∥ quadrilateral BEDF is parallelogram, (7 points) BD and EF were equally divided
RELATED INFORMATIONS
- 1. In ▱ ABCD, equilateral △ ade and equilateral △ BCF are made inward with AD and BC as sides respectively, connecting be and DF. It is proved that the quadrilateral BEDF is a parallelogram
- 2. In the parallelogram ABCD, take ad and BC as sides, make positive △ ade and positive △ BFC outward respectively, connect dB and EF at point O, and prove that the quadrilateral debf is a parallelogram
- 3. Given the parallelogram ABCD, take ad BC as the edge, make positive △ ade and positive △ BCF outside the parallelogram, connect BD and EF, and they intersect at O, and prove EO = fo, do = Bo There is no picture
- 4. Given the parallelogram ABCD, make the equilateral triangle ade and equilateral triangle BCF outward with AD and BC, connect be and DF, and find be = DF
- 5. The two diagonals of quadrilateral ABCD are perpendicular to each other and intersect at O. given AC = 4cm, BD = 5cm, find the area of quadrilateral ABCD
- 6. The length of diagonal lines AC and BD of quadrilateral ABCD are m and N respectively. It can be proved that when AC ⊥ BD (as shown in Figure 1), the area of quadrilateral ABCD is s = 12mn. Then when the acute angle between AC and BD is θ (as shown in Figure 2), the area of quadrilateral ABCD is s = () A. 12mnB. 12mnsinθC. 12mncosθD. 12mntanθ
- 7. If the area of parallelogram ABCD is 100, then s △ PAB + s △ PCD =
- 8. It is known that the area of parallelogram ABCD is s, and the area of triangle PAB and triangle PCD are S1 and S2 respectively If point P is outside the parallelogram ABCD, then S1 + S2 -- half s Please explain why
- 9. As shown in the figure, in the pyramid s-abcd, the bottom surface ABCD is a parallelogram, the side SBC ⊥ the bottom surface ABCD, ∠ ABC = 45 °, SA = sb, proving that SA ⊥ BC
- 10. Given any parallelogram ABCD, e is the midpoint of AD, f is the midpoint of BC, the proof is: ab + DC = 2ef
- 11. As shown in the figure, ABCD is a parallelogram, EF is parallel to ac. if the area of the triangle is 4 square centimeters, calculate the area of the triangle CDF?
- 12. In the parallelogram ABCD, make a straight line AF through a, cross BC and E, and cross DC extension line to F. it shows that the area of triangle ABF and ade is equal
- 13. In the parallelogram ABCD, a straight line AF passing through point a intersects point BC and point E, and a DC extension line intersects point F, which indicates that the areas of △ ABF and △ ade are equal thinking Yes
- 14. In a parallelogram ABCD, a straight line AF passing through point a intersects point BC at point E, and a DC extension line intersects point F. try to explain that the area of △ ABF and △ ade is equal
- 15. In the parallelogram ABCD, BC = 2Ab, e is the midpoint of BC, then ∠ AED=______ ．
- 16. In the parallelogram ABCD, BC = 2Ab, e is the midpoint of BC, then ∠ AED=______ ．
- 17. In the parallelogram ABCD, BC = 2Ab, e is the midpoint of BC, then ∠ AED=______ ．
- 18. ）(1) As shown in Figure 10, in the parallelogram ABCD, BC = 2Ab, e is the midpoint of the BC side, and the degree of ∠ AED is calculated. (2) as shown in Figure 11, e is the square ab ）(1) As shown in Figure 10, in the parallelogram ABCD, BC = 2Ab, e is the midpoint of BC side, and the degree of ∠ AED is calculated (2) As shown in Figure 11, e is a point in the square ABCD, and △ Abe is an equilateral triangle. Think about the relationship between ∠ CED and ∠ CEB, and explain the reason (3) As shown in Figure 12, the height and perimeter of the isosceles trapezoid ABCD are determined by the upper bottom ad = 1, the lower bottom BC = 3 and the diagonal AC ⊥ BD file:///C:/Documents%20and%20Settings/Administrator/Local%20Settings/Temporary%20Internet%20Files/ Content.IE5/3YP2UMUP/image035%5B1%5D .jpg
- 19. In the parallelogram ABCD, BC = 2Ab, e is the midpoint of BC, then ∠ AED=______ ．
- 20. The area of trapezoidal ABCD is 45 square centimeters, and its height is 6 centimeters. AC and BD intersect at point e. the area of AED is 5 square centimeters,