![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
In the parallelogram ABCD, M is the midpoint of BC, n is on AB, BN = 1 / 2An, NM intersects BD with G, and BG: GD is obtained
Let NB = 1, then an = 2, DC = AB = 3, triangle nbg is equal to triangle MCE, so CE = NB = 1, de = DC + CE = 3 + 1 = 4, triangle nbg is similar to triangle DEG (the reason is found), GB: GD = NB: de = 1:4
In the parallelogram ABCD, BC has a point E, be: EC = 3:2, f is the midpoint of AB, EF intersects BD with G, and BG: GD is obtained
Suppose that AC and BD intersect at O, BG: Bo = 1: X, ABO area = 1 / 2 * AB * Bo * cosabo, BFG area = 1 / 2 * BF * BG * cosabo = 1 / 2 * (1 / 2 * AB) * (1 / X * Bo) = 1 / (2x) ABO area = 1 / (4x) ABC area; similarly, we can get BCG area = 3 / (10x) ABC area, and EFB area = BFG area + BCG area = 11 / (20x) ABC area = 1 / 2 * 3 / 5abc area, But it's all the same, similar to calculation
As shown in the figure, points E and F are on the extension line of edge ab of ▱ ABCD, and be = AB, BF = BD. connecting CE and DF intersects at point M. are CD and cm equal? Please give reasons
(1 point) reason: AB is parallel and equal to CD from ABCD, then be is parallel and equal to CD if ∠ 2 = ∠ f ① (2 points) and be = AB, so becd is parallelogram. (4 points) then BD ‖ CE, so ∠ 1 = ∠ 3 ② (5 points) and BD = BF, so ∠ 1 = ∠ f ③ (6 points) from ①, ② and ③, so CD = cm (8 points)
The parallelogram ABCD, take a point be = AB, BF = BD on the extension line of AB, connect CE, DF and intersect m, indicating that CD = cm
For BF = BD
Then BFD = BDF
BF is parallel to CD
Then BFD = FDC
Angle DMC = angle BDM (i.e. angle BDF)
So angle DMC = angle FDC
If the two base angles are equal, they are isosceles triangles, so cm = CD
It is known that in the parallelogram ABCD, e, F, G and H are points on AB, BC, CD and Da respectively, and AE = CG, BF = DH
It is proved that the ∵ quadrilateral ABCD is a parallelogram, a = C, ad = BC, ∵ BF = DH, ∵ ah = CF, ∵ in △ AEH and △ CGF, ah = CF, a = CAE = CG, ≌ AEH ≌ CGF (SAS)
In the parallelogram ABCD, AB > ad, AE, BF, CG and DH are the bisectors of the internal angles
AB > ad in parallelogram ABCD,
AE, BF, CG and DH are bisectors of internal angles,
CD, AB, e, F, G, h, DH and AE, respectively,
CG crossed with P, m, BF and AE, CG crossed with N, G,
Verification: ab = AD + PQ
Parallel + bisector = isosceles triangle
The triangle ADH is the waist triangle, ad = ah
Similarly, BC = BG = ah = ad
The congruence of APH and gqb can be obtained
Then pH and GB are parallel and equal
PQ=HB
AB=AD+PQ=AH+HB
As shown in the figure, it is known that the quadrilateral ABCD is a parallelogram. On the extension line of AB, intercept be = AB, BF = BD, connect CE and DF, and intersect at point M. verification: CD = cm
It is proved that: ∵ quadrilateral ABCD is a parallelogram, ∵ AB is parallel and equal to DC. And ∵ be = AB, ∵ be is parallel and equal to DC. The ∵ quadrilateral bdce is a parallelogram. ∵ DC ∵ BF, ∵ CDF = ∵ F. similarly, ∵ BDM = ∵ DMC. ∵ BD = BF, ∵ BDF = ∵ f.
RELATED INFORMATIONS
- 1. In the trapezoidal ABCD, point P starts from point a and moves along the edge of ad to point d at a speed of 1cm / s, and point Q starts from point C and moves along CB to point B at a speed of 2cm / s. if points P and Q start from two points at the same time, when one of them reaches the end point, the other stops moving. (1) what is the value of T when trapezoidal pbqd is a parallelogram? (2) What is the value of T when trapezoid pbqd is isosceles trapezoid?
- 2. The perimeter of parallelogram ABCD is 75cm, the height is 14cm with BC as the bottom (as shown in the figure); the height is 16cm with CD as the bottom. Calculate the area of parallelogram ABCD
- 3. As shown in the figure below, the area of rectangle ABCD is 12 square decimeters, so the area of circle is () square decimeters Well, there is no picture. Let's give a brief description There is a circle, no matter how big, half of it (that is, a semicircle). Draw a rectangle on the semicircle. The length of the rectangle is equal to the diameter of the circle. Then say that the area of the rectangle is 12 square decimeters. Tell the area of the whole circle. Thank you!
- 4. In the quadrilateral ABCD, BD bisects ∠ ABC, AC ⊥ BD at R, PQ intersects BC and ad at points Q and P respectively, and ∠ bad = ∠ BQP
- 5. As shown in the figure, ad ∥ BC in ladder ABCD, point E is the midpoint of edge ad, connecting be with AC at point F, the extension line of be with CD at point G, verify Ge / GB = AE / BC
- 6. As shown in the figure, in ladder ABCD, ab ‖ CD, e is the midpoint of DC, the line be intersects AC at F, and the extension line of ad intersects g; verification: EF · BG = BF · eg
- 7. Take a point G on the extension line of one side of the square ABCD and connect it with a, intersect BC at point F, connect diagonal BD and intersect AF at point E. AE = 5, AF = 3 are known, and calculate the length of FG chart
- 8. In quadrilateral ABCD, ad = BC, AE, CF are perpendicular to BD, be = DF. It is proved that quadrilateral ABCD is parallelogram I can't find the last condition in the verification. Please give me some advice. Thank you
- 9. In square ABCD, point G starts from point C and moves along CB; point h starts from point a and moves along Ba, and two points move at the same speed and stop at the same time
- 10. In the parallelogram ABCD, there is a point E on the side of DC, which connects AE. The point F is on the side of AE, and connects B. It is proved that the triangle ABF is similar to the triangle EAD Angle BFE is equal to angle C
- 11. In the parallelogram ABCD, be ⊥ ad, BF ⊥ CD, the perpendicular foot is e, F, the perpendicular center of △ bef is h, if DG ⊥ BC, the perpendicular foot is g It should be verification: BH = GF
- 12. In trapezoidal ABCD, AB / / CD, AC is vertical to BD, angle BDC = 60 degrees, AC = 3 root sign 3, find the length of BD and the area of trapezoidal ABCD In the isosceles trapezoid ABCD, ab = 2CD, AC bisects DAB, ab = 4, ab ‖ CD, (1) Find the trapezoid angle; (2) find the trapezoid area
- 13. In a circle with a radius of 12 decimeters, draw a square, draw a small circle in the square, and calculate the area of the small circle
- 14. In the diamond ABCD with side length of 2, ∠ bad = 60 °, e is the midpoint of AB, P is the upper moving point of AC, and the minimum value of Pb + PE is obtained
- 15. In rectangular ABCD, BC = 4, BG is perpendicular to diagonal AC and intersects AC, ad and ray CD at points e, F, G, ab = x respectively. (1) when point G coincides with point D, calculate the value of X; (2) when point F is the midpoint of AD, calculate the value of X and the sine value of ∠ ECF
- 16. As shown in the figure, points E and F are points on edge BC and CD of diamond ABCD respectively, and ∠ EAF = ∠ d = 60 ° and ∠ fad = 45 °, then ∠ CFE=______ Degree
- 17. As shown in the figure, in diamond ABCD, ∠ bad = 60 °, M is the midpoint of AB, P is a moving point on the diagonal AC, if the minimum value of PM + Pb is 3, then the length of AB is () A. 3B. 3C. 6D. 23
- 18. It is known that the four vertices of the quadrilateral ABCD are a (2,3), B (1. - 1), C (- 1. - 2) and D (- 2.2),
- 19. If the length of a diagonal line of diamond ABCD is 6 sides and the length of AB is a root of the square of equation x - 7x + 12 = 0, then the perimeter of diamond ABCD is In the two known equations x ^ 2 + ax + (A-1) = 0, the absolute value of negative root is greater than that of positive root, and the value range of a? It is known that the equation x ^ 2-mx + M-1 = 0 has two real roots x1, X2 and 1 / X1 + 1 / x2 = m, then M =? One diagonal of the diamond ABCD is 3
- 20. In diamond ABCD, if the angle ABC is equal to 120 degrees, then BD is equal to several times AC