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ABCD is trapezoid, ab ∥ CD, ADBE is parallelogram, the extension line of AB intersects EC in F Can s △ BCE be one third of s ladder ABCD? If not, explain the reason; if yes, find out the relationship between AB and CD The answer is yes. How can we prove it
Let be be intersect CD with G, and let the trapezoid height be h, because be = ad = BG, the distance from e to DC is 2h, so s △ BCE = s △ ECG - △ BCG = CG * 2H / 2-cg * H / 2 = CG * H / 2, and s ladder ABCD = (AB + CD) * H / 2, so 3cg * H / 2 = (AB + CD) * H / 2, and CG = cd-ab, and simplify the substitution to get DC = 2Ab, so when DC = 2Ab, it is full
As shown in the figure, the quadrilateral ABCD is an isosceles trapezoid, and the ADBE & nbsp; is a parallelogram with an area equal to 8. We also know that the area of the triangle BCE is 2, so what is the area of the triangle CDE?
Because ab ∥ CD, the △ foe is similar to △ CDE, the similarity ratio is 1:2, so EF: EC = 1:2. Because the area of △ BCE is 2, the area of △ bef is 2 × 12 = 1. In parallelogram ADBE, the area of △ BOE is 14 × parallelogram ADBE = 14 × 8 = 2, so the area of △ foe is 1 + 2 = 3 The area of △ CDE is 3 × 4 = 12. Method 2: because ADBE is a parallelogram, so s △ Abe = s △ abd = s △ ade = s △ BDE = 4. Because quadrilateral ABCD is isosceles trapezoid, so ab ‖ CD. So the distance from point B to CD is half of the distance from e to CD. So s △ CDE = 2 (s △ abd + s △ BCE) = 2 × (4 + 2) = 12 The area is 12
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