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As shown in the figure: given rectangular ABCD, AC and BD intersect at O, AE ‖ BD, de ‖ AC. verification: OE ⊥ ad
It is proved that: ∵ AC, BD are diagonals of rectangles, ∵ AC = BD, Ao = do, ∵ AE ∥ BD, de ∥ AC, ∵ quadrilateral Aode is parallelogram, ∵ Ao = do, ∵ quadrilateral Aode is diamond, ∵ OE ⊥ ad
As shown in the figure, in rectangular ABCD, the bisector of ∠ bad intersects BC at point E, and point O is the intersection of diagonals, and ∠ CAE = 15 °, then ∠ BOE=______ Degree
As shown in the figure, connecting OE; ∵ quadrilateral ABCD is a rectangle, and EA bisects ∵ bad, ∵ BAE = 45 °; ∵ Abe is an isosceles right triangle, ab = be; ∵ CAE = 15 °, ∵ Bao = ∵ CAE + ∵ BAE = 60 °; and ∵ OA = ob, ∵ △ Bao is an equilateral triangle, ab = Bo; ∵ Bo = be; ∵ OBC = 90 ° - ∵ ABO = 30 °; ∵ BOE = (180 ° - 30 °) △ 2 = 75 °. So the answer is 75
As shown in the figure, in rectangular ABCD, AC and BD intersect at point O, AE bisects ∠ bad, BC intersects at E. if ∠ EAO = 15 °, then the degree of ∠ BOE is______ Degree
In rectangular ABCD, ∵ AE bisects ∵ bad, ∵ BAE = ∵ ead = 45 ° and it is also known that ∵ EAO = 15 ° and ∵ OAB = 60 ° and ∵ OA = ob, ∵ boa is equilateral triangle, ∵ Ba = Bo, ∵ BAE = 45 ° and ∵ ABC = 90 ° and ∵ Bae is isosceles right triangle, ∵ Ba = be. ∵ be = Bo, ∵ EBO = 30 ° and ∵ BOE = ? BeO. At this time, ? BOE = 75 °
As shown in the figure, in rectangular ABCD, the diagonal AC and BD intersect at point O. is there a circle so that all four points of ABCD are on the circle? If so, please point out the center and radius of the circle; if not, please explain the reason
Existence is a circle with o as the center and 1 / 2 diagonal as the radius (OA)
RELATED INFORMATIONS
- 1. As shown in the figure, the quadrilateral ABCD is inscribed in the circle, the diagonal AC and BD intersect at the point E, f is on AC, ab = ad, ∠ BFC = ∠ bad = 2 ∠ DFC. Verification: (1) CD ⊥ DF; (2) BC = 2CD
- 2. Given that ad is parallel to BC and E is the midpoint of Cd in trapezoidal ABCD, is the area of trapezoidal ABCD twice that of triangular Abe
- 3. As shown in the figure, in the trapezoidal ABCD, ab ‖ CD is known, and point E is the midpoint of BC. If the area of △ DEA is S1 and the area of trapezoidal ABCD is S2, then the quantitative relationship between S1 and S2 is______ ．
- 4. In diamond ABCD, AD / / BC, ab = CD, e is the midpoint of the bottom BC, connecting AE and de I'm sorry. I didn't pay attention to it It's in ABCD
- 5. The diagonal lines AC and BD of isosceles trapezoid ABCD intersect at the two bottoms of O, OE parallel trapezoid, and the area of △ BOC is 16 square centimeter It's urgent. I'm begging you. Please
- 6. As shown in the figure, in ▱ ABCD, e and F are two points on AC, be ∥ DF
- 7. In quadrilateral ABCD, AC and BD are perpendicular to each other, AC = 18 cm, BD = 13 cm. Find the area of quadrilateral ABCD
- 8. As shown in the figure, the diagonal lines AC and BD of quadrilateral are perpendicular to each other, AC + BD = 10. When the length of AC and BD is what, the area of quadrilateral ABCD is the largest?
- 9. As shown in the figure, P is a point on ▱ ABCD. Given s △ ABP = 3 and s △ PCD = 1, the area of parallelogram ABCD is () A. 6B. 8C. 10d. Not sure
- 10. In the parallelogram ABCD, CE ⊥ CD is passed through point C, ad is crossed with point E, and the segment EC is rotated 90 degrees counterclockwise around point e to obtain the segment EF
- 11. As shown in the figure, the diagonals AC and BD of rectangle ABCD intersect at O, AE ‖ BD, de ‖ AC
- 12. As shown in the figure, in isosceles trapezoid ABCD, ad ‖ BC. AB = DC, e is the midpoint of BC, connecting AE and De, proving: AE = De
- 13. In the parallelogram ABCD, AE ⊥ BC in E, AF ⊥ CD in F, ∠ BAE = 30 °, be = 2, CF = 1, (1) find the area of triangle ECD, (2) Find the length of EG
- 14. It is known that e is the edge of the parallelogram ABCD, AB extends to the upper point, de intersects BC with F. proof: the area of the triangle ABF is equal to the area of the triangle efc Mathematics in grade two
- 15. In the quadrilateral ABCD, the diagonal AC bisects ∠ bad, AB > ad
- 16. ABCD is a parallelogram, e and F are the midpoint of AD and BC respectively
- 17. 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
- 18. 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
- 19. 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
- 20. 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