It is known that in the quadrilateral ABCD as shown in the figure, AD / / BC, e is the midpoint of ab. it is proved that s quadrilateral ABCD = 2S triangle CDE

It is known that in the quadrilateral ABCD as shown in the figure, AD / / BC, e is the midpoint of ab. it is proved that s quadrilateral ABCD = 2S triangle CDE

Make the height of parallelogram through e, and the perpendicular foot is f
Then: EF is the height of parallelogram ABCD and triangle CDE
Area of parallelogram: S = CD * ef
Area of triangle: S = CD * EF / 2
So

As shown in the figure, the quadrilateral ABCD is a square with side length of 2, and E is a point on the side of AD. rotate △ CDE counterclockwise around point C to △ CBF, connect EF and BC to point g. if EC = eg, then De=______ ．

The ∵ CDE rotates counterclockwise around point C to ∵ CBF, ∵ DCE = ∵ BCF, CE = CF, de = BF, ∵ square ABCD, ∵ DCB = 90 °, CD = ad = AB = BC = 2, ∵ ECB + ∵ BCF = 90 °, ∵ CEF is isosceles right triangle, ∵ Efec = 2, ∵ EC = eg, ∵ ECG = ∵ EGC = ∵ BGF, ∵ D

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

Because the area ratio of triangle ABC to triangle ade is 3:4, so AB: de = 3:4, then AB: CE = 3:1, because triangle ABF is similar to triangle FCE, and the similarity ratio is 3:1, then their area ratio is 9:1, 9 + 1 = 10, so the sum of triangle ABF and triangle FCE area is: 10 △ 810 = 12

The area ratio of triangle ABC to triangle ade is 3:4, and the area of triangle ABF is 10 square centimeters larger than that of triangle FCE
Finding the area of parallelogram ABCD

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