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
Δ ABC is an equilateral triangle. ∵ CE = CD, ∵ d = ∵ Dec, ∵ ECB = ∵ D + ∵ Dec = 2 ∵ D. ∵ be = De, ∵ EBC = ∵ D. ∵ ECB = 2 ∵ EBC. Also ? be ⊥ CE, ? ECB = 60 °. ? be ⊥ CE, AE = CE, ? AB = BC
RELATED INFORMATIONS
- 1. 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
- 2. D. E and F are the midpoint of AB, BC and AC respectively. It is proved that triangle ABC is similar to triangle EFD
- 3. 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
- 4. 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
- 5. In △ ABC, points D.E.F are AB.BC.CA To prove that △ ABC is similar to △ EFD
- 6. There is a 15cm * 13.5cm * 4.5cm box. How long can you put a chopstick? What about a box of size a * b * C? Trigonometric function has not yet learned, with the most basic Pythagorean theorem
- 7. Pythagorean theorem in junior high school If the sides of a right triangle are a, B and the hypotenuse is C, then the relationship between a, B and C can be expressed as
- 8. A question about Pythagorean theorem in junior high school? A door frame (ABCD) is 2m long and 1m wide. Can a 3 m long and 2.2m wide veneer pass through the door frame? Why?
- 9. Junior high school inverse proportion function + Pythagorean theorem We know that n is a positive integer, P1 (x1, Y1), P2 (X2, Y2) ,P(xn,yn),… Is the inverse scale function y = K / x, where X1 = 1, X2 = 2 ,xn=n,… Let A1 = x1y1, A2 = x2y2 ,An=xn yn+1,… If A1 = a (a is a nonzero constant), then A1 · A2 · ·The value of an is_______ (expressed by an algebraic expression containing a and N)
- 10. Pythagorean theorem and all the formulas of geometry Geometric formulas and conversions are all in urgent need
- 11. 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
- 12. 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
- 13. The triangle ABC is an equilateral triangle, CE ⊥ plane ABC, BD ∥ CE, and CE = CA = 2bd, M is the midpoint of EA
- 14. As shown in the figure, in △ ABC, ab = AC, and points D, e and F are the midpoint of △ ABC
- 15. As shown in the figure, in △ ABC, ab = AC, and points D, e and F are the midpoint of △ ABC
- 16. As shown in the figure, in △ ABC, ab = AC, and points D, e and F are the midpoint of △ ABC
- 17. 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
- 18. 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______ .
- 19. 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
- 20. 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