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
It is proved that triangle ABF has the same base and height as parallelogram ABCD, so its area is half of parallelogram ABCD
Same: Triangle ade is the same as parallelogram ABCD, so its area is half of parallelogram ABCD
So the area of △ ABF is equal to that of △ ade
RELATED INFORMATIONS
- 1. 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
- 2. 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
- 3. 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?
- 4. It is known that, as shown in the figure, the quadrilateral ABCD is a parallelogram, and both △ ade and △ BCF are equilateral triangles
- 5. 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
- 6. 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
- 7. 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
- 8. 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
- 9. 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
- 10. 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
- 11. In the parallelogram ABCD, BC = 2Ab, e is the midpoint of BC, then ∠ AED=______ .
- 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. )(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
- 15. In the parallelogram ABCD, BC = 2Ab, e is the midpoint of BC, then ∠ AED=______ .
- 16. 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,
- 17. 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
- 18. 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
- 19. 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
- 20. What is the ratio of s △ AED to s ladder ABCD if ab ‖ DC and point e are the midpoint of waist BC in ladder ABCD