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As shown in the figure, in the parallelogram ABCD, points E and F are on CD and ab respectively, and DF / / BF is on the point
A quadrilateral ABCD is a parallelogram,
∴AB∥CD,
∵DF∥BE,
The quadrilateral BEDF is a parallelogram,
∴OE=OF．
As shown in the figure, the diagonal lines AC and BD of parallelogram ABCD intersect at point O, and the straight line EF passes through point O and intersects at AB and CD respectively
Verification: quadrilateral aecf is parallelogram.
The quantitative relationship between the perimeter or area of the quadrilateral afed and the quadrilateral BFEC is explained
It is proved that: (1) because ABCD is a parallelogram, so Ao = Co, AB / / DC, so angle EAC = angle FCA, angle AEF = angle CFE, so triangle AOE is all equal to triangle COF, (angle, angle, edge) so EO = fo, because Ao = Co, EO = fo, so quadrilateral aecf is a parallelogram (four sides of diagonal line bisecting each other
As shown in figure 16.1.36, it is known that the diagonal lines AC and BD of the parallelogram ABCD intersect at point O, and make a straight line EF through o to intersect AB and CD at points E and f respectively
Verification: OE = of
Proof: link BF, De
Because ABCD is a parallelogram,
So AB / / CD, Ao = Co,
So EAO = FCO, AEO = CFO,
So triangle AEO is equal to triangle CFO (a, a, s)
So OE = of
In the isosceles triangle ABCD, e is the midpoint of CD, EF is perpendicular to AB and F, if AB = 6, EF = 5, calculate the area of trapezoid
If AE and be are connected, the area of trapezoid is equal to the sum of triangle ade, AEB and EBC. Because point E is the midpoint of De, the heights of triangle ade and EBC are equal to 1 / 2 of the height of trapezoid, and the sum of the heights of triangle ade and triangle ECB is the height of trapezoid. Let the height of trapezoid be H, Trapezoidal area can also be expressed as (AD + BC) * H / 2, and 1 / 2 * AB * EF + 1 / 2 * ad * H / 2 + 1 / 2 * BC * H / 2 = (AD + BC) * H / 21 / 2 * 6 * 5 + 1 / 2 * (AD + BC) * H / 2 = (AD + BC) * H / 215 = 1 / 2 * (AD + BC) * H / 2 (AD + BC) * H / 2 = 30 is trapezoidal area
RELATED INFORMATIONS
- 1. As shown in the figure, fold a square piece of paper ABCD along EF, AD / / BC. If the angle EFG = 52 °, calculate the degree of angle deg
- 2. As shown in the figure, after a rectangular piece of paper ABCD is folded along EF, points D and C respectively fall at d 'C', the intersection of ED 'and BC is g, if ∠ EFG = 65 degrees Find the degree of ∠ 1 and ∠ 2
- 3. When a rectangular piece of paper ABCD is folded along EF, the intersection points of ED and BC are g, D and C at M and N, respectively. If ∠ EFG = 55 °, then ∠ 1=______ °，∠2=______ °．
- 4. (Mianyang, 2011) as shown in the figure, fold a rectangular piece of paper ABCD with a length of 8cm and a width of 4cm so that point a and point C coincide, and the length of crease EF is equal to 2525cm
- 5. The edge length of rectangular paper ABCD is ab = 4, ad = 2. Fold the rectangular paper along EF so that point a and point C coincide. After folding, color it on one side (as shown in the figure), the area of the colored part is () A. 8B. 112C. 4D. 52
- 6. In the parallelogram ABCD, ab = KBC, point P is on the diagonal BD, and angle ∠ ape = ∠ bad, PE and BC are compared with point E, and the quantitative relationship between PA and PE is explored
- 7. As shown in the figure, in the parallelogram ABCD, e is the midpoint of AD, the intersection of the extension line of CE and Ba and the point F, if BD = 2CD, the verification: ∠ f = ∠ BCF
- 8. In the parallelogram ABCD, is the midpoint of e-type CD, a point on the edge of F-type BC, and the angle FAE = angle ead, EF and AE perpendicular? Give reasons
- 9. In the parallelogram ABCD, what is the length of CE if the extension line of AE bisector angle bad and intersecting DC is at point E, AB equals 4 and ad equals 9?
- 10. As shown in the figure, in the parallelogram ABCD, e and F are two points on the diagonal BD, and be = DF, connecting AE, AF, CE and cf. what kind of quadrilateral is the quadrilateral aecf, which shows your truth
- 11. As shown in the figure, it is known that in the parallelogram ABCD, e and F are respectively the midpoint of AD, G and H are two points on the diagonal BD, and BG = DH, what conclusion can be obtained?
- 12. Parallelogram ABCD, AC = 2gc, AE = EF = FB, triangle GEF area is 6 square centimeter, find the area of parallelogram
- 13. The parallelogram ABCD, e, f are the trisection points of BC, connecting BD, AE, AF to intersect BD with m, N, finding: Mn: Nd It is known that: the parallelogram ABCD, f is the trisection point of BC, connecting BD, AE, AF with BD in M, and finding: Mn: Nd
- 14. It is known that the perimeter of the parallelogram ABCD is 32, ab = 4, then BC is equal to
- 15. As shown in the figure, the diagonals AC and BD of rectangular ABCD are called o, and E F are the midpoint of OA and ob respectively If ad = 4cm, ab = 8cm, find the length of DF of EF
- 16. As shown in the figure, the diagonals AC and BD of rectangle ABCD intersect at O, e and F are the midpoint of OA and ob respectively. (1) prove: △ ade ≌ △ BCF; (2) if ad = 4cm, ab = 8cm, find the length of CF
- 17. If vector AB = vector DC, then ABCD is a parallelogram? Is that right? Maybe it's not collinear
- 18. It is known that in the quadrilateral ABCD, ab = ad = 4, BC = 6, CD = 2,3 vector AB * vector AD + 4 vector CB * vector CD = 0, we can find the circumcircle radius r of triangle ABC
- 19. In rectangle ABCD, ab = 2, ad = 1, then vector AC × vector CD
- 20. In the diamond ABCD with side length of 1, ∠ bad = 60 ° and E is the midpoint of BC, then the vector AE is multiplied by the vector AC