It is known that as shown in the figure, ∠ ACD is an external angle of △ ABC, CE and CF divide ∠ ACB, ∠ ACD, eh ‖ BC equally, and intersect AC and CF at points g and h respectively. Verification: Ge = GH
∵EH∥BC,∴∠BCE=∠GEC,∠GHC=∠DCH,∵∠GCE=∠BCE,∠GCH=∠DCH,∴∠GEC=∠GCE,∠GCH=∠GHC,∴EG=GC=GH,∴GE=GH.
RELATED INFORMATIONS
- 1. CE and CF are the bisector of the inner angle and the bisector of the outer angle of the triangle ABC, respectively
- 2. As shown in the figure, in △ ABC, ∠ a = 70 ° bisector CE ‖ ab of outer angle, calculate the degree of ∠ B and ∠ ACB
- 3. As shown in the figure, in the triangle ABC, ∠ a = 40 °, CE is the bisector of the outer angle of ∠ ABC, and CE ‖ AB, calculate the degree of ∠ B and ∠ ACB
- 4. Let a = (COS α, sin β) and B = (COS β, sin α), then α - β =? 0
- 5. Given the constant a > 0, the straight line passing through the fixed point a (0, a), with m vector = (λ, a) as the direction vector and passing through the fixed point B (0, a), and with n vector = (1,2 λ, a) Where λ belongs to R, the equation for finding the locus C of point P is obtained
- 6. Let a = (1, e ^ - x) and B = (e ^ x, m), where m is a constant and m ∈ R. f (x) = a · B When m = - 1, find the inequality f (x ^ 2-3) + F (2x)
- 7. It is known that a is a constant, and a > 0, vector M = (√ x, - 1), vector n = (1, ln (x + a)), Finding the maximum value of the function f (x) = m · n in the interval (0,1]
- 8. a. B is a nonzero vector. "A ⊥ B" is "function f (x) = (XA + b) · (xb-a) is a linear function It is a necessary and insufficient condition,
- 9. Let a and B be non-zero vectors. If f (x) = (XA + b) (a-xb), X belongs to R and is even, A. A vertical B B. a parallel B C. | a | = | B | D. | a | unequal | B|
- 10. Given that a and B are nonzero vectors and the function f (x) = (XA + b) (a-xb), a necessary and insufficient condition for the image of F (x) to be a parabola symmetric about the y-axis is A.a⊥b B. A parallel B C.|a|=|b| D.a=b
- 11. As shown in the figure, ∠ ACD = 60 ° and ∠ BCD = 40 °, CE and CF are the bisectors of △ ACD and △ BCD respectively, and the degree of ∠ ECF can be calculated
- 12. In the triangle ABC, the bisector of the angle ABC intersects the bisector of the angle ACD at e, and the angle a is equal to 78 degrees
- 13. As shown in the figure, in △ ABC, a = 70 ° and B = 50 °, CD bisects ACB, and calculates the degree of ACD
- 14. In the quadrilateral ABCD, BC = 2, DC = 4, and ∠ A: ∠ ABC: ∠ C: ∠ ADC = 3:7:4:10, find the length of ab
- 15. As shown in the figure, C is a point on the line AB, △ ACD and △ BCE are equilateral triangles. AE intersects CD at point m, BD intersects CE at point n, AE intersects AE at point O. The results are as follows: (1) AOB = 120 °; (2) cm = cn; (3) Mn ‖ ab
- 16. It is known that C is a point on the line AB, and △ ACD and △ BCE are equilateral triangles
- 17. In the quadrilateral ABCD, ab = 2, radical 5, BC = 2, CD = 1, ad = 5, and angle c = 90 degrees, find the area of quadrilateral ABCD
- 18. E is a point in the square ABCD, and the angle ade is equal to the angle DAE is equal to 15 degrees. It is proved that BCE is an equilateral triangle
- 19. E is a point in the square ABCD, and there is ∠ ade = ∠ DAE = 15 ° to prove that △ BCE is an equilateral triangle
- 20. E is a point in the square ABCD, and △ EAB is an equilateral triangle, then ∠ ade=______ .