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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
To connect AC, \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\howdo you like it If △ ACF (ASA), ｜ AE = AF, and ∠ EAF = ∠ d = 60 °, then △ AEF is an equilateral triangle, ｜ AFE = 60 °, and ∠ AFD = 180 ° - 45 ° - 60 ° = 75 °, then ∠ CFE = 180 ° - 75 ° - 60 ° = 45 °. So the answer is: 45
As shown in the figure, in the trapezoidal ABCD, ad ∥ BC, ab = CD, BC = 6, ad = 3, point m is the midpoint of edge BC, with m as the vertex, EMF = angle B, ray me intersects AB at point E
Trapezoid ABCD, AD / / BC, ab = CD = BC = 6, ad = 3, M is the midpoint of BC, with m as the vertex, make ∠ EMF = ∠ B, ray me intersects AB and E, ray MF intersects CD in F, connecting EF
If the triangle FEM is isosceles triangle, find the length of EB
In △ BEM, ∠ B + ∠ BEM = ∠ EMC = ∠ EMF + ∠ FMC
From the known ﹥ B = ﹥ EMF, ﹥ BEM = ﹥ FMC is obtained
In trapezoidal ABCD, ab = CD, so ∠ B = ∠ C, plus the proved ∠ BEM = ∠ FMC
There are △ BEM ∽ CMF
Since △ EFM is an isosceles triangle, we discuss it in three small cases
1° EM=FM
&If △ BEM ≌ CMF is obtained by adding △ BEM ≌ CMF, then be = cm
&And M is the midpoint of BC, CM = BC / 2 = 6 / 2 = 3, so be = cm = 3
2 ° & nbsp; EF = FM, then ∠ EMF = ∠ FEM
&From the proved △ BEM ∽ CMF, be / cm = EM / MF
&Since m is the midpoint of BC, BM = cm and EF = FM, be / BM = EM / EF
&By adding ∠ B = ∠ EMF = ∠ FEM, △ EBM ∽ MEF can be obtained
&Nbsp; & nbsp; so ∠ BEM = ∠ EMF = ∠ B, then BM = em
&Through M for Mg ⊥ be in G, through a for ah ⊥ BM in H
&In the isosceles triangle BEM, BM = em, BG = be / 2
&In the isosceles trapezoid ABCD, ab = CD, BH = (BC-AD) / 2 = (6-3) / 2 = 3 / 2
&From ∠ AHB = ∠ MGB = 90 ° and common angle ∠ B = ∠ B, △ ABH ∽ MBG is obtained
&Then AB / MB = BH / BG, that is BG = BH * MB / ab
&Since BG = be / 2, BH = 3 / 2, MB = BC / 2 = 6 / 2 = 3, ab = 6
&So be / 2 = 3 / 2 * 3 / 6 = 3 / 4, then be = 2 * 3 / 4 = 3 / 2
3° EF=EM
&There are two kinds of pictures of small cases together, in which the letter with "&" 39; in the upper right corner indicates the third kind of small case
&This is done to take advantage of symmetry:
&Nbsp; & nbsp; in the third case, E & # 39; F & # 39; = E & # 39; m is the EF = FM in the second case
&So in the third case, E & # 39; B & # 39; is CF in the second case
&Nbsp; & nbsp; then return to the previous small case and calculate CF
&According to the conclusion of the previous small case, be = 3 / 2
&Nbsp; & nbsp; just prove △ BEM ∽ CMF, there is be / cm = BM / CF, that is CF = BM * cm / be
&Since BM = cm = BC / 2 = 6 / 2 = 3, CF = 3 * 3 / (3 / 2) = 6
&In the second case, be = 6
To sum up, be = 3 / 2 or 3 or 6
(be = 6 when EF = em; be = 3 / 2 when Fe = FM; be = 3 when me = MF)
In the parallelogram ABCD, CE ⊥ ad, CF ⊥ ad, e, f are perpendicular feet, CE = CF, indicating that the parallelogram is a diamond
The question is wrong!
A point C outside the straight line ad cannot have two perpendicular lines perpendicular to the known straight line!
In parallelogram ABCD, CE ⊥ ad CF ⊥ AB description ∠ FCE = ∠ D
This is the problem on page 58 of the third new mathematics in the eighth grade
There's a picture, but it can't come up
I hope someone can help me
The triangle GFC is similar to GCB The rest is not difficult I didn't give you a cent
It is known that, as shown in the figure, the diagonals of rectangle ABCD intersect at O, de ‖ AC, AE ‖ dB, and AE and de intersect at E
It is proved that: ∵ de ∥ AC, AE ∥ dB, ∵ quadrilateral Aode is parallelogram, ∵ rectangle ABCD diagonal intersects at O, ∵ Ao = do, ∵ quadrilateral doae is diamond
It is known that, as shown in the figure, the diagonals of rectangle ABCD intersect at O, de ‖ AC, AE ‖ dB, and AE and de intersect at E
It is proved that: ∵ de ∥ AC, AE ∥ dB, ∵ quadrilateral Aode is parallelogram, ∵ rectangle ABCD diagonal intersects at O, ∵ Ao = do, ∵ quadrilateral doae is diamond
It is known that, as shown in the figure, the diagonals of rectangle ABCD intersect at O, de ‖ AC, AE ‖ dB, and AE and de intersect at E
It is proved that: ∵ de ∥ AC, AE ∥ dB, ∵ quadrilateral Aode is parallelogram, ∵ rectangle ABCD diagonal intersects at O, ∵ Ao = do, ∵ quadrilateral doae is diamond
As shown in the figure, the diagonals of rectangle ABCD intersect at o.de ‖ AC, CE ‖ BD. (1) prove that quadrilateral OCED is diamond; (2) if ∠ ACB = 30 °, the area of diamond OCED is 83, calculate the length of AC
(1) It is proved that: the ∵ de ∥ OC, CE ∥ OD, ∵ quadrilateral OCED is parallelogram, ∵ quadrilateral ABCD is rectangle, ∵ Ao = OC = Bo = OD, ∵ quadrilateral OCED is diamond; (2) ∵ ACB = 30 degree, ∵ DCO = 90-30 degree = 60 degree
The diagonals of rectangle ABCD intersect at O, de ∥ AC, CE ∥ BD. if the angle ACB = 30 degrees, the area of rhombic OCED is 8 root sign 3, the length of AC is calculated
∵ ABCD is a rectangle, ∵ ABC = 90 °, ACB = 30 °, ∵ BC = √ 3AB, ∵ area of ABC = (1 / 2) BC × AB = (√ 3 / 2) AB ^ 2, ∵ area of BCD = (√ 3 / 2) AB ^ 2. ∵ ABCD is a rectangle, ∵ Bo = do, ∵ area of OCD = (1 / 2) △ area of BCD = (√ 3 / 4) AB ^ 2
If ad = 3, BC = 9, then go: BG=______ ．
Let do = x, then Bo = 3x, BG = GD = 2x, go = 2x-x = x, go: BG = x: 2x = 1:2
RELATED INFORMATIONS
- 1. 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
- 2. 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
- 3. 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
- 4. 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
- 5. 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
- 6. 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
- 7. 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?
- 8. 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
- 9. 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!
- 10. 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
- 11. 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
- 12. 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),
- 13. 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
- 14. In diamond ABCD, if the angle ABC is equal to 120 degrees, then BD is equal to several times AC
- 15. If the central line and the high line on the hypotenuse of a right triangle are 6cm and 5cm respectively, then its area is
- 16. In rectangle ABCD, ab = 4cm, ad = 3cm, fold the rectangle along the straight line AC, and point B falls at point E, connecting de. what is aced in four sides? Please explain the reason and calculate its area! Very urgent, this is today's home work, tomorrow to hand in, online 30 minutes, please hurry up, good answer bonus!
- 17. M. What is the relationship between the area of Mn parallel to AC, △ ABM and △ BCN?
- 18. As shown in the figure, △ ABC has rhombic ampn, if ammb = 12, then BPBC=______ ．
- 19. As shown in the figure, △ ABC, Ag ⊥ BC is at point G, AB and AC are taken as one side to make rectangle ABME and rectangle acnf outside △ ABC respectively, and the ray GA intersects EF at point h. If AB = KAE, AC = Kaf, try to explore the quantitative relationship between he and HF, and explain the reason
- 20. As shown in the figure, ab = ad, BC = De, BA ⊥ AC, Da ⊥ AE, can you prove am = an?