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______ ．

Take the midpoint F of AD, connect EF, make DM ⊥ AB and m through D, cross EF to N, ∵ trapezoidal ABCD, DC ∥ AB, e is the midpoint of BC, f is the midpoint of AD, ∥ EF ∥ ab ∥ CD, EF = 12 (AB + CD), ∵ DM ⊥ AB, ∥ DM ⊥ EF, ∥ S1 = 12ef × DN + 12ef × Mn = 12ef × DM, S2 = 12 (CD + AB) × DM = EF × DM, ∥ S2

As shown in the figure on the right, in the trapezoid ABCD, CD and ab are the top and bottom of the trapezoid respectively. AC and BD intersect at point E, and let the area of triangle ade be S1 and triangle BCE
The area of S2 is ()
A、S1＜S2 B、S1＝S2 C、S1＞S2

B
If the area of triangles with the same base and height is equal, - S (ADE) = s (BCE)
Then subtract the common part of the triangle Dec from both sides at the same time,
That is: S1 = S2

It is known that: as shown in the figure, DC ‖ AB, AC = BC, angle ACB = 90 °, BD = AB, AC and BD intersect at e in ladder ABCD. It is proved that △ ade is an isosceles triangle

Let BC = AC = 1, then AB = √ 2, so BD = √ 2
In △ BCD, ∠ BCD = 135 ° and sin ∠ BDC = 1 / 2 is obtained from sine theorem, so ∠ BDC = 30 °
From DC ∥ AB, there is ∠ abd = ∠ BDC = 30 °, thus ∠ ADB = 75 ° and ∠ AED = ∠ abd + ∠ BAE = 75 °
So ∠ ADB = ∠ AED, ad = AE, △ ade is isosceles triangle

As shown in the figure, it is known that a, B, C and D are four points on ⊙ o, and that DC and ab intersect at point E. if BC = be, it is proved that △ ade is an isosceles triangle

It is proved that ∵ a, D, C and B are four points in common circle, ∵ BC = be, ∵ BCE = e, ∵ a = e, ∵ ad = De, i.e. △ ade is isosceles triangle

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______ ．