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

Connecting BD and AF at point G, triangle abd and triangle ABF have the same area (same base and equal height). Because triangle ABG and triangle GED have the same area (same base and equal height), triangle ABF and triangle ade have the same area

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1. 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
2. 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?
3. It is known that, as shown in the figure, the quadrilateral ABCD is a parallelogram, and both △ ade and △ BCF are equilateral triangles
4. 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
5. 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
6. 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
7. 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
8. 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
9. 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
10. If the area of parallelogram ABCD is 100, then s △ PAB + s △ PCD =
11. 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
12. In the parallelogram ABCD, BC = 2Ab, e is the midpoint of BC, then ∠ AED=______ ．
13. In the parallelogram ABCD, BC = 2Ab, e is the midpoint of BC, then ∠ AED=______ ．
14. In the parallelogram ABCD, BC = 2Ab, e is the midpoint of BC, then ∠ AED=______ ．
15. ）(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
16. In the parallelogram ABCD, BC = 2Ab, e is the midpoint of BC, then ∠ AED=______ ．
17. 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,
18. In ladder ABCD, ad ‖ BC.AC Intersection with BD at point E, if the area of △ AED is a, △ BEC area B, calculate the area of trapezoid ABCD
19. In diamond ABCD, point E is on diagonal AC, point F is on the extension of BC, EF = EB, EF and CD intersect at point G How to prove that triangle EGC is similar to triangle DGF after connecting DF
20. As shown in the figure, in ladder ABCD, ab ‖ DC, ab ⊥ BC, e are the midpoint of AD, AB + BC + CD = 6, be = 5, then the area of ladder ABCD is equal to () A. 13B. 8C. 132D. 4