In the trapezoidal ABCD, AD / / BC, e is the midpoint of waist AB, de ⊥ CE

In the trapezoidal ABCD, AD / / BC, e is the midpoint of waist AB, de ⊥ CE

It is proved that extending CB de intersection point M
∵AD//BC
∴∠A=∠ABM
∵∠AEM=∠MEB AE=EB
∴△AED≌△BEM
∴DE=EM MB=AD
∵DE⊥CE
∴CM=DC
∴AD+BC=CD

As shown in Figure 1, in the right angle trapezoid ABCD, ad ‖ BC, ab ⊥ BC, ∠ DCB = 75 ° and the other vertex e of the equilateral triangle DCE with CD on one side is on ab
F is a point on CD, ∠ FBC = 30?: DF = FC

Connect AC, because triangle DCE is equilateral triangle, so de = DC = CE, ∠ Dec = ∠ EDC = ∠ ECD = 60 ° because ad ∥ BC, so, ∠ ADC +, ∠ BCD = 180 °, ∠ bad +, ∠ ABC = 180 ° because ∠ DCB = 75 ° so ∠ ADC = 105 ° so ∠ ade = 45 ° because ab ⊥ BC, so ∠ B = 90 ° so ∠ bad = 90 °