Taking the two right sides AB and BC of the right triangle ABC as one side, make the equilateral triangle △ Abe and equilateral △ BCF outward respectively, connecting EF and EC. Try to explain: (1) EF = EC; (2) EB ⊥ CF
This paper proves that: all the \ \ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\and ∵ EF = EC, ∴EB⊥CF.
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- 1. Known: as shown in the figure, take two right angle sides AB and BC of RT △ ABC as sides to make equilateral △ Abe and equilateral △ BCF, connect EF and EC, please explain EF = EC
- 2. In the triangle ABC, CD is high, e is the midpoint of CD, the extension line of AE intersects CB at point F, AE = CE and AF is vertical to CB. If EF = 1, then AE
- 3. As shown in the figure, in RT △ ABC, ∠ B = 90 °, ab = 1, BC = 12, take point C as the center, CB as the radius of arc intersection Ca at point D; take point a as the center, ad as the radius of arc intersection AB at point E. (1) find the length of AE; (2) take points a and E as the center, AB length as the radius of arc, two arcs intersect at point F (F and C are on both sides of AB), connect AF and EF, let the circle where EF intersects de at point G, connect AG, try to guess Think about the size of EAG and explain the reason
- 4. In triangle ABC, D is the point on AC, passing through point D as de parallel to CB, intersecting point AB at point E, and the area ratio of triangle ade to triangle ace is 1:16
- 5. As shown in the figure, the side length of equilateral triangle ABC is 6, ad is the middle line on the side of BC, M is the moving point on ad, e is the point on AC, if AE = 2, then the minimum value of EM + cm is?
- 6. Let the area of the quadrilateral ABCD be one, divide the edge ad into three equal parts, the dividing point is E.F, let AE = EF = FD, and divide BC into three equal parts, the dividing point is H.G, so that BH = Hg = GC Connecting EF and GH to prove s quadrilateral efgh = 1 / 3
- 7. As shown in the figure, the side length of equilateral △ ABC is 6, ad is the middle line on BC side, M is the moving point on ad, e is a point on AC side. If AE = 2, the minimum value of EM + cm is () A. 27B. 4C. 37D. 1+27
- 8. As shown in the figure: the side length of equilateral △ ABC is 5, ad is the middle line on BC side, M is the moving point on ad, e is a point on AC side. If AE = 2, the minimum value of EM + cm is________ The title says that the length of the side is 5. I can't find it on the Internet. I want the mid-term exam tomorrow
- 9. As shown in the figure, the side length of equilateral △ ABC is 12, ad is the middle line on BC side, M is the moving point on ad, e is a point on AC side, if AE = 4, the minimum value of EM + cm is______ .
- 10. As shown in the figure, in △ ABC, ab = AC, be = AE, the perimeter of △ BCE is 12, BC = 5, find the length of ab
- 11. Taking the two right sides AB and BC of the right triangle ABC as one side, make the equilateral triangle △ Abe and equilateral △ BCF outward respectively, connecting EF and EC. Try to explain: (1) EF = EC; (2) EB ⊥ CF
- 12. As shown in the figure, △ Abe and △ ACD are respectively formed by △ ABC folding along AB and AC sides. If ∠ BAC = 150 °, then ∠ DAB =, ∠ DAE= Why
- 13. It is known that: as shown in the figure, in RT △ ABC, ∠ C = 90 ° and ∠ a = 30 ° take AB and AC as edges respectively, and make equilateral △ Abe and equilateral △ ACD on the outside of △ ABC, de and ab intersect with F, proving: EF = FD
- 14. What is the 20th term of the arithmetic sequence 8,5,2
- 15. Find 2,5,8,11 The 33rd term of this arithmetic sequence The formula of addition, subtraction, multiplication and division
- 16. The process of finding the 20th term of arithmetic sequence 8,5,2, - 1
- 17. Arithmetic sequence 8, 5, 2 Item 20 of the report is______ .
- 18. From 1, 2, 3 How many different arithmetic sequences are there if you take any three different numbers from the 20 numbers of, 20 to form the arithmetic sequence
- 19. Insert three numbers between 20 and 70, so that these five numbers become arithmetic sequence. What are the three numbers inserted?
- 20. Will 1, 2 If the nine numbers are divided into three groups averagely, the probability that the three numbers in each group can form an arithmetic sequence is () A. 156B. 170C. 1336D. 1420