As shown in the figure, in △ ABC, ab = AC, and points D, e and F are the midpoint of △ ABC
It is proved that: ∵ D, e and F are the midpoint of △ ABC, respectively, ∵ de ∥ 12ac, EF ∥ 12ab, ∵ quadrilateral ADEF is a parallelogram, and ∵ AC = AB, ∵ de = EF. & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ∵ quadrilateral ADEF is a diamond
RELATED INFORMATIONS
- 1. As shown in the figure, in △ ABC, ab = AC, and points D, e and F are the midpoint of △ ABC
- 2. As shown in the figure, in △ ABC, ab = AC, and points D, e and F are the midpoint of △ ABC
- 3. The triangle ABC is an equilateral triangle, CE ⊥ plane ABC, BD ∥ CE, and CE = CA = 2bd, M is the midpoint of EA
- 4. As shown in the figure, △ ABC is an equilateral triangle, BD ⊥ plane ABC, BD ∥ CE, and CE = CA = 2bd, M is the midpoint of EA, find plane DEA ⊥ plane ECA
- 5. It is known that the triangle ABC is an equilateral triangle, EC is perpendicular to the plane ABC, BD is perpendicular to the plane ABC, and EC and DB are on the same side of the plane ABC It is known that the triangle ABC is an equilateral triangle, EC is perpendicular to plane ABC, BD is perpendicular to plane ABC, and EC and DB are on the same side of plane ABC, M is the midpoint of EA, CE = 2bd. It is proved that: (1) plane BDM is perpendicular to plane ECA; (2) plane DEA is perpendicular to plane ECA
- 6. As shown in the figure: in △ EBD, EB = ed, point C is on BD, CE = CD, be ⊥ CE, a is a point on the CE extension line, EA = EC. Try to judge the shape of △ ABC and prove your conclusion
- 7. As shown in the figure: in △ EBD, EB = ed, point C is on BD, CE = CD, be ⊥ CE, a is a point on the CE extension line, EA = EC. Try to judge the shape of △ ABC and prove your conclusion
- 8. D. E and F are the midpoint of AB, BC and AC respectively. It is proved that triangle ABC is similar to triangle EFD
- 9. As shown in the figure, ACB = BDC = CED = EFD = 90 degree. (1) how many triangles in the figure are similar to △ ABC. (2) which two triangles in the figure are similar to each other
- 10. As shown in the figure, in the triangle ABC, BC = AB, f is a point on the side of AC, e and D are points on BC and ab respectively, ∠ EFD = a Prove the CEF ∽ AFD of triangle
- 11. Given that the points D, e and F are respectively the midpoint of the edges AB, BC and Ca of △ ABC, connecting de and EF, if we want to make the quadrilateral ADEF diamond, we need to add the following conditions______ Just fill in one
- 12. In the figure, AE: EC = 1:2, CD: DB = 1:4, BF: FA = 1:3, and the area of △ ABC is s = 1, then the area of quadrilateral AFHG is______ .
- 13. As shown in the figure, in △ ABC, ad: DB = 2:1, be: EC = 3:1, CP: FA = 4:1, then △ dep is a fraction of the area of △ ABC
- 14. Ad is the height of triangle ABC, AB is equal to 10, ad is equal to 8, BC is equal to 12, triangle ABCD is isosceles triangle
- 15. In the triangle ABC, AB is equal to 13, BC = 10, and the middle line ad on the side of BC = 12. Is the triangle ABC an isosceles triangle
- 16. As shown in the figure, it is known that ad, be and CF are respectively the heights of the three sides of △ ABC, h is the perpendicular, and the extension line of ad intersects the circumscribed circle of △ ABC at point G
- 17. As shown in the figure, it is known that ad, be and CF are respectively the heights of the three sides of △ ABC, h is the perpendicular, and the extension line of ad intersects the circumscribed circle of △ ABC at point G
- 18. As shown in the figure, ⊿ ABC is an isosceles triangle, ∠ ABC = 90? Therefore, B = 10, D is a point outside ⊿ ABC, connecting AD.BD If point h, AC ⊥ abd is an equilateral triangle, find the length de: (2) if BD = AB, and DH BH = 3 / 4
- 19. In the triangle ABC, BD is perpendicular to AC at point D, CE is perpendicular to ab at point E, BD and CE intersect at point h, ad = DH = 1, CD = 5, find the area of triangle ABC. If you can't send the graph, please draw it by yourself. The top vertex is a, and the bottom is left B and right C
- 20. The relationship between EF and be can be obtained by taking m as a 30 ° angle, intersection AB with E, intersection CA extension line with F