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As shown in the figure, in the quadrilateral ABCD, the diagonal AC and BD intersect at point E, and ∠ cab = ∠ CBD. Given AB = 4, AC = 6, BC = 5, de = 5.5, find the length of de. & nbsp; & nbsp; [please solve this problem from the Perspective of similar triangles,]
In ▱ ABCD, AC is a diagonal, B = ∠ CAD, extend BC to point E, make CE = BC, connect de. (1) prove: quadrilateral abed is isosceles trapezoid; (2) if AB = ad = 4, find the area of trapezoid abed
(1) It is proved that: in ▱ ABCD, ad ∥ BC, ab = CD, ∥ CAD = ∥ ACB. ∥ B = ∥ CAD, ∥ ACB = ∥ B. ∥ AB = AC. ∥ ab ∥ CD, ∥ B = ∥ DCE. And ∥ BC = CE, ≌ ABC ≌ DCE (SAS). ∥ AC = de = ab. ∥ ad ∥ be, ∥ quadrilateral abed is isosceles trapezoid
As shown in the figure, in the parallelogram ABCD, AC is diagonal, e is on DC, de: CE = 1:2, find a
Because AE bisects ∠ a DF bisects angle D, so angle DAE = angle BAE, angle CDF = angle ADF, because it is a parallelogram, so ad is parallel to BC, so angle DAE = angle AEB, angle ADF = angle DFE (the parallel inner stagger angles of two lines are equal) because angle DAE = angle BAE, angle CDF = angle ADF, so angle BAE = angle AEB, angle CDF = angle DFE (equivalent substitution)
In the parallelogram ABCD, e and F are the points on the edge of AD and ab respectively, and be = DF, be and DF intersect at point G
It is proved that CN ⊥ be, CH ⊥ DF are made by C respectively, connecting CE and CF, ∵ s △ BCE = 12s, parallelogram ABCD = s △ DFC, ∵ 12 · DF · ch = 12 · be · CN, ∵ be = DF, ∵ CN = ch, ∵ GC bisection ∠ bgd (the point with equal distance to both sides of the angle is on the bisection line of the angle)
As shown in the figure, in the parallelogram ABCD, e and F are the points on the sides of AD and AB, be = DF, be and DF intersect with G, and the bisection angle bgd of GC is proved
In parallelogram ABCD, the area of △ BCE and △ CDF is half of parallelogram ABCD. Why?
In the parallelogram ABCD, it is easy to find that the area of △ BCE and △ CDF is half of that of parallelogram ABCD. Why? Parallelogram ABCD and △ BCE have the same bottom and height (bottom BC) sabcd = bottom * height s △ BCE = bottom * height / 2S △ BCE / sabcd = (bottom * height / 2) / (bottom * height) = 1 / 2. The area of △ BCE is parallelogram ABC
In the parallelogram ABCD, e and F are the points on the edge of AD and ab respectively, and be = DF, be and DF intersect at point G
It is proved that CN ⊥ be, CH ⊥ DF are made by C respectively, connecting CE and CF, ∵ s △ BCE = 12s, parallelogram ABCD = s △ DFC, ∵ 12 · DF · ch = 12 · be · CN, ∵ be = DF, ∵ CN = ch, ∵ GC bisection ∠ bgd (the point with equal distance to both sides of the angle is on the bisection line of the angle)
As shown in the figure, in the quadrilateral ABCD, points E and F are two points on the diagonal BD, and be = DF (1) if the quadrilateral aecf is a parallelogram, prove that ABCD is also a parallelogram. (2) if the quadrilateral aecf is a diamond, prove that the quadrilateral ABCD is also a diamond
It is proved that if AC is connected, let AC and BD intersect at point O. (1) ∵ quadrilateral aecf is a parallelogram, ∵ OE = of, OA = OC, ∵ be = FD, ∵ ob = OD. ∵ quadrilateral ABCD is a parallelogram. (2) ∵ quadrilateral aecf is a diamond, ∵ OE = of, OA = OC, AC ⊥ BD. ∵ be = FD, ∵ ob = OD. ∵ quadrilateral ABCD is a diamond
The diamond ABCD points E and F are on the diagonal BD, be = DF = 1 / 4bd. If the quadrilateral aecf is a square, find sin ∠ Abe
∵ square aecf let AE = a ∵ AE = AF = CE = CF, AC and EF be equally divided with each other ∠ AEC = 90 °, AC = √ (AE & # 178; + CE & # 178;) = √ 2 × a ∵ EF and AC be equally divided with each other ∵ Ao = OC = √ 2 / 2 × a ∵ OE = Ao = √ 2 / 2 × a ∵ be = DF = 1 / 4bd ∵ ob = 2oe = √ 2 × a ∵ AB = √ (AO & ∵ 178; + ob & ∵ 178;) = √
In diamond ABCD, be = DF, it is proved that quadrilateral aecf is diamond
It is proved that AC and BD are diagonals of square ABCD
AC and BD are divided vertically and equally,
∴AE=EC,AF=FC,
∵BE=DF,∴DE=BF,
∴OE=OF
∵EC= OE2+OC2,FC= OC2+OF2,
∴EC=FC,
AE = EC = AF = FC
The quadrilateral ABCD is a diamond
It is known that in the triangle ABC, D is the point on BC, e.f.h.g is AC.CD.DB The midpoint of AB, EF + ad = 6cm, find the length of GH
Urgent --- urgent --- urgent -!
We know that EF is the median of △ CAD, EF = 1 / 2ad,
Similarly, GH is the median of △ bad, GH = 1 / 2ad
So 1 / 2ad + ad = 6cm, ad = 4cm,
GH=2cm
RELATED INFORMATIONS
- 1. As shown in Figure 1, in quadrilateral ABCD, ad ∥ BC, de bisects ∥ ADB, ∥ BDC = ∥ BCD, (1) (2) if the bisector of abd and the extension of CD intersect at f (Fig. 2), and F = 55 °, find ABC
- 2. It is known that as shown in the figure, ad and BC intersect at point O, ∠ cab = ∠ DBA, AC = BD
- 3. The quadrilateral ABCD is the inner boundary quadrilateral of the circle, the diagonal AC and BD intersect at e, the extended DA and CB intersect at F, the angle CAD = 60 degrees, DC = De, it is proved that a is the outer center of the triangle bef How does angle F + angle AEB equal to 60 degrees?
- 4. In the isosceles triangle ABCD, ab = AC, ∠ a = 36 ° and BD is the square of ∠ ABC, then calculate AD / AC
- 5. As shown in the figure, in the square ABCD, point P is a point on the diagonal AC, PE ⊥ AB, PF ⊥ BC, and the perpendicular feet are points E and f respectively. If the perimeter of the square ABCD is 8cm, then the perimeter of the quadrilateral EBFP is______ cm．
- 6. It is known that in trapezoidal ABCD, ad is parallel to BC, AB is parallel to de, AF is parallel to DC, e and F are on edge BC As shown in the figure, in the trapezoidal ABCD, ad ‖ BC, ab ‖ De, AF ‖ DC, e and F are on the side BC, and the quadrilateral aefd is a parallelogram (1) What is the equivalent relationship between AD and BC (2) When AB = DC, it is proved that the quadrilateral aefd is a rectangle I want to prove that the process is not an equation
- 7. As shown in the figure, in trapezoidal ABCD, ad is parallel to BC, AF is vertical, BC and points F, m are the midpoint of CD, angle B is 45 degrees, AF is 4, EF is 7 Find ABCD area
- 8. In the trapezoidal ABCD, ab = CD, e is the midpoint of AD
- 9. As shown in the figure, in trapezoidal ABCD, ∠ d = 90 ° and M is the midpoint of ab. if cm = 6.5 and BC + CD + Da = 17, the area of trapezoidal ABCD is () A. 20B. 30C. 40D. 50
- 10. In trapezoidal ABCD, ad ∥ BC, ab = 4, BC = 9, CD = 5, Da = 6. (1) prove ab ⊥ BC; (2) find the area of trapezoidal ABCD
- 11. AB = AC = ad, ∠ DAC = 3 ∠ cab Angle BDC = 1 / 2 angle cab, angle DBC = 1 / 2 angle DAC How did you get this step?
- 12. Isosceles trapezoid ABCD, ab = CD = 5, ad = 4, BC = 10, e, f on AB, BC, EF bisects the circumference of trapezoid, BF = x, and uses X to express the area of △ bef AB DC is waist
- 13. As shown in the figure, in square ABCD, point m is a point on diagonal BD. through point m, make me parallel to CD, intersect BC at point E, and make MF parallel to BC at point F. prove am = EF As shown in the figure, in the square ABCD, point m is a point on the diagonal BD, passing through point m to make a point Me parallel CD intersects BC at point E, MF parallel BC at point F. prove am = EF
- 14. As shown in the figure, m and N are the midpoint of AD and DC on both sides of the square ABCD, cm and BN intersect at P
- 15. As shown in the figure, in the quadrilateral ABCD, ab = CD, points E and F are the midpoint of AD and BC respectively, GH ⊥ EF intersects point P
- 16. As shown in the figure, in ladder ABCD, ab ‖ DC and E are the midpoint of BC, and the extension lines of AE and DC intersect at point F to connect AC and BF. What quadrilateral is abfc? Please give reasons
- 17. As shown in the figure, the quadrilateral ABCD is a square, AB is an equilateral triangle with an edge to the square, Abe, CE and DB intersect at point F, then ∠ AFD=______ Degree
- 18. Given that the length of a diagonal line of a diamond is 4cm and the perimeter is 16cm, what are the degrees of the four inner angles of the diamond?
- 19. If the perimeter of the diamond is 8cm and the height is 1cm, the angle of the diamond is smaller______ ．
- 20. As shown in the figure, the parallelogram ABCD, e is the midpoint of AD, AC and be intersect at the point F, s △ EFC = 1, then s parallelogram ABCD=