As shown in the figure, in ladder ABCD, AB is parallel to DC, point E is the midpoint of DC, ∠ AED = ∠ BEC, proving that ladder ABCD is isosceles ladder

Because AB is parallel to DC
Therefore, AED = EAB and EBA = bec
Because ∠ AED = ∠ bec
Therefore, EAB = EBA
So the triangle EAB is an isosceles triangle, EA = EB
So the triangle AED is congruent with the triangle bec
So ad = BC
So trapezoid ABCD is isosceles trapezoid

RELATED INFORMATIONS

1. In rectangular trapezoid ABCD, ab ∥ CD, ab ⊥ ad, and ab = 13, CD = 8, ad = 12, then the distance from point a to BC is______ ．
2. In the right angle trapezoid ABCD, ab ∥ CD, Da is vertical AB, ab = 13, CD = 8, ad = 12, then what is the distance from point a to BC?
3. As shown in the figure, in the rectangular trapezoid ABCD, the bottom AB = 13, CD = 8, ad ⊥ AB and ad = 12, then the distance from a to BC is () A. 12B. 13C. 12×2113D. 10.5
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5. As shown in the figure, in rectangular trapezoid ABCD, ab ∥ CD, ad ⊥ DC, ab = BC, and AE ⊥ BC. (1) prove: ad = AE; (2) if ad = 8, DC = 4, find the length of ab
6. As shown in the figure, in rectangular trapezoid ABCD, ab ∥ CD, ad ⊥ DC, ab = BC, and AE ⊥ BC. (1) prove: ad = AE; (2) if ad = 8, DC = 4, find the length of ab
7. In rectangular trapezoid ABCD, ab ∥ CD, ad ⊥ DC, ab = BC, and AE ⊥ BC
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11. As shown in the figure, in trapezoidal ABCD, ∠ d = 90 ° and M is the midpoint of ab. if cm = 6.5 and BC + CD + Da = 17, the area of trapezoidal ABCD is () A. 20B. 30C. 40D. 50
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14. As shown in the figure, ladder ABCD, e is the midpoint of AB, DC = AD + BC, prove de ⊥ EC
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