In the trapezoidal ABCD, ab = CD, e is the midpoint of AD

In the trapezoidal ABCD, ab = CD, e is the midpoint of AD


AB = CD, isosceles trapezoid, angle a = angle D, e is the midpoint of AD, AE = De, triangle Abe and triangle DCE are congruent, EB = EC



In the trapezoidal ABCD, ab ‖ CD, ∠ a = 90 °, ab = 2, BC = 3, CD = 1, e is the midpoint of AD, and ∠ BEC = 90 ° is proved


Certification:
Extend CE and cross the extension line of BA at point F
∵AB‖CD
∴∠DCF=∠F,∠D=∠FAE
∵AE=DE
∴△AEF≌△CED
∴EF=CE,AF=CD=1
∴BF=2+1=3
∴BF=BC
∵EF=EC
∴BE⊥CF
That is, BEC = 90 degree



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


Extend the intersection point of CM and DA at E. ∵ ad ∥ BC, ∵ Mae = ∥ B, ∥ e = ∥ BCM. Am = BM, ≌ ame ≌ BMC. ∥ me = MC = 6.5, AE = BC. BC + CD + Da = 17, ∥ d = 90 °, de + DC = 17 ①, de2 + DC2 = CE2 = 169 ②. ∥ de · CD = 12 [(de + DC) 2-de2-dc2] = 60. ∥ the area of trapezoidal ABCD is 12de · CD = 30