Let E and f be on the edge BC and CD of rectangular ABCD respectively, and the areas of △ Abe, △ ECF and △ FDA are a, B and C respectively. Find the area s of △ AEF

Let AB = x1, be = X2, EC = X3, CF = x4, then FD = x1-x4, ad = x2 + X3. From the meaning of the title, we get x1 ·x2 = 2A, x3 ·x4 = 2B, (x1-x4) × (x2 + x3) = 2c, that is, x1 ·x3-x2 ·x4 = 2 (B + C-A), and then substitute x1x2x3x4 = 4AB into x2x4 = x1x3-2 (B + C-A) to get the univariate quadratic equation about x1x3

RELATED INFORMATIONS

1. Rectangle ABCD, e and F are on BC and CD respectively. The area of triangle Abe, CEF and ADF are 2, 3 and 4 respectively
2. As shown in the figure, ABCD is a rectangle, e and F are the points on BC and CD respectively, and s △ Abe = 3, s △ CEF = 2, s △ ADF = 2, then s △ AEF = ()
3. As shown in the figure, in square ABCD, e and F are on BC and CD, respectively, ∠ EAF = 45 ° to prove s △ AEF = s △ Abe + s △ ADF
4. As shown in the figure, in square ABCD, e and F are on BC and CD, respectively, ∠ EAF = 45 ° to prove s △ AEF = s △ Abe + s △ ADF
5. In square ABCD, e and F are on BC and CD, respectively, ∠ EAF = 45 °, try to explain s △ AEF = s △ Abe + s △ ADF
6. It is known that the quadrilateral ABCD is a square with a side length of 6 cm. The area of the triangle ECF is 3 square cm larger than that of the triangle ADF
7. As shown in the figure, it is known that the quadrilateral ABCD is a square with a side length of 5cm. The area of the triangle ECF is 5 square centimeters larger than that of the triangle ADF. Find the length of the line CE
8. If the areas of △ CEF, △ Abe and △ ADF are 3, 4 and 5 respectively, the area of △ AEF can be obtained
9. The quadrilateral ABCD is a square, and △ AEF is an equilateral triangle, where the points E and F are on BC and CD respectively. The detailed explanation of s △ CEF = s △ Abe + s △ ADF is obtained The area of s △ Abe and s △ ADF is equal to half ab. s △ CEF = half (a-b) &# 178; after that, how can we do it? We can't do it any more. Let's do it as soon as possible
10. Given that ab = 8cm, ad = 12cm, the area of triangle Abe and triangle ADF are respectively 13% of rectangle ABCD, the area of triangle AEF is calculated
11. As shown in the figure, in rectangular ABCD, e is the point on BC and F is the point on CD. Given s △ Abe = s △ ADF = 13sabcd, then the value of s △ AEF: s △ CEF is equal to () A. 2B. 3C. 4D. 5
12. In the isosceles trapezoid ABCD, ab ∥ DC, ad = BC = 5, DC = 7, ab = 13, point P starts from point a and moves to the midpoint C along ad → DC at the speed of 2 unit lengths per second At the same time, point Q starts from point B and moves along Ba to terminal a at the speed of 1 unit length per second. Let the movement time be T seconds (1) When the value of T is, the quadrilateral pqbc is a parallelogram (2) In the whole movement process, when t is the value, the quadrilateral with points B, C, P and Q as the vertex is isosceles trapezoid
13. Cut AE = BF = CG = DH on each side of square ABCD, connect AF, BG, CH and De, and intersect a ` B ` c ` d 'in turn. It is proved that quadrilateral a ` B ` c ` d' is a square
14. It is known that the upper middle points of AB and BC on both sides of triangle ABC are E.F. two points on AC are G.H. Ag = GH = HC,; Connect and extend EG.FH It is proved that ABCD is a parallelogram;
15. In △ ABC, E.F AB.CB G. h is fed to two points on AC, and Ag = GH = HC EG.FH Intersection at point D. proof: a quadrilateral ABCD is a parallelogram
16. As shown in the figure, in square ABCD, e and F are two points on BD, and be = DF
17. The quadrilateral ABCD is a diamond. The extension of de perpendicular to ab intersects BA at point E, and DF perpendicular to BC intersects BC at point F. please guess the size of De and DF
18. 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
19. 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
20. 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