Given the square ABCD, the straight line passing through C intersects the extension lines of AD and ab at points E and f respectively, and AE = 15 and AF = 10, the side length of square ABCD is calculated
∵ BC ∥ AE ∥ FBC ∥ FAE ∥ bcae = fbfA if the side length of a square is x, then x15 = 10 − X10 ∥ x = 6
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- 1. Given the square ABCD, the straight line passing through C intersects the extension lines of AD and ab at points E and f respectively, and AE = 15 and AF = 10, the side length of square ABCD is calculated
- 2. Given the square ABCD, the straight line passing through C intersects the extension lines of AD and ab at points E and f respectively, and AE = 15 and AF = 10, the side length of square ABCD is calculated
- 3. Given that the straight line of square ABCD passing through point C intersects the extension lines of AD and ab with points E and f respectively, and AE = 10 and AF = 15, the side length of square ABCD is calculated
- 4. In square ABCD, e is the point on AC, EF ⊥ AB, eg ⊥ ad, ab = 6, AE: EC = 2:1. Find the area of quadrilateral afeg
- 5. In square ABCD, e is the point on AC, EF ⊥ AB, eg ⊥ ad, ab = 6, AE: EC = 2:1. Find the area of quadrilateral afeg
- 6. In square ABCD, e is on AC, EF ⊥ AB is on F, eg ⊥ ad is on G, ab = 8cm, AE ratio EC = 3:1, calculate the area of quadrilateral afeg Draw your own picture I can't make it easy
- 7. In square ABCD, e is a point on AC, EF is perpendicular to AB, BG is perpendicular to ad, ab = 6, AE: EC = 2:1, find the area of afeg
- 8. In the quadrilateral ABCD, point E is the midpoint of BC, point F is the midpoint of CD, and AE is perpendicular to AB, AF ⊥ CD, connecting EF to prove AB = ad When AB, CD and BC are related, the △ AEF is an equilateral triangle
- 9. In square ABCD, point F is on ad, point E is on AB, AE = EB, AF = 1 / 4AD Connecting CE, EF (Pythagorean theorem, inverse theorem)
- 10. In the square ABCD, the midpoint e of BC connects AE, and a point F on CD makes EF ⊥ AE connect AF. verify that AF = AD + FC
- 11. Please read the following materials: problem: as shown in Figure 1, in diamond ABCD and diamond befg, points a, B and E are on the same line, P is the midpoint of line DF, connecting PG and PC. if ∠ ABC = ∠ bef = 60 °, explore the position relationship between PG and PC and the value of pgpc. XiaoCong's idea is: extend GP intersection DC at point h, construct congruent triangle, and solve the problem through reasoning. Please refer to XiaoCong Students' ideas, explore and solve the following problems: (1) write the position relationship between PG and PC and the value of pgpc in the above problem; (2) rotate the diamond BFG in Figure 1 clockwise around point B, so that the diagonal BF of diamond BFG is in the same line with the edge ab of diamond ABCD, and other conditions in the original problem remain unchanged (as shown in Figure 2). Are the two conclusions you get in (1) correct What's the change? Write down your conjecture and prove it. (3) if ∠ ABC = ∠ bef = 2 α (0 °< α< 90 °) in Figure 1, rotate the diamond BFG clockwise at any angle around point B, and other conditions in the original problem remain unchanged, please write down the value of pgpc directly (expressed by the formula containing α)
- 12. Can the four vertices of diamond ABCD be on the same circle? If they are on the same circle, what shape should it be
- 13. As shown in the figure, the area ratio of triangle ABC to triangle ade is 3:4, and the area of triangle ABF is 10 square centimeter larger than that of triangle FCE
- 14. In trapezoidal ABCD, ab ∩ CD, CE bisection ∩ BCD intersects point E, and CE ⊥ ad, de = 2ae, if s △ CDE = 1, then the area of quadrilateral abce is
- 15. In trapezoidal ABCD, ab ‖ CD, CE bisects ∠ BCD, and CE ⊥ ad is at point E, de = 2ae, the area of △ CDE is 8, then the area of trapezoidal ABCD is () A.16 B.15 C.14 D.13 The answer seems to be B, but please tell me why
- 16. As shown in the figure below, the quadrilateral ABCD is trapezoid, the ratio of the top and bottom is 3:5, and E is the midpoint of the ad side. Calculate the area ratio of the triangle CDE and the quadrilateral abce
- 17. A flat quadrilateral ABCD is divided into a triangle CDE and a trapezoid abce by the line CE, with the height of AF = 6cm. It is known that the area of trapezoid abce is 9cm larger than that of triangle CDE. How many cm is the length of trapezoid AE?
- 18. As shown in the figure, the quadrilateral ABCD and the quadrilateral defg are both square. The area of the triangle AFH is known to be 6 square centimeters, so the area of the triangle CDH can be calculated
- 19. The point m is the midpoint of AB at the bottom of the isosceles trapezoid ABCD
- 20. As shown in the figure, in the parallelogram ABCD, ad = 2Ab, point m is the midpoint of AD, and the degree of ∠ BMC is calculated