In diamond ABCD, AD / / BC, ab = CD, e is the midpoint of the bottom BC, connecting AE and de I'm sorry. I didn't pay attention to it It's in ABCD

In diamond ABCD, AD / / BC, ab = CD, e is the midpoint of the bottom BC, connecting AE and de I'm sorry. I didn't pay attention to it It's in ABCD

Trapezoidal ABCD, ab = CD
Then ABCD is isosceles trapezoid
Then angle ABC = angle DCB
It's the midpoint of the bottom BC again
Then be = CE
And ab = CD
Then {Abe ≌} DCE (edge)
So AE = De
So triangle ade is isosceles triangle
Get proof

As shown in the figure, e is the point on the waist BC of the right angle trapezoid ABCD, ab ‖ DC, angle B = 90 °, and ∠ AED = 90 ° to find ab × DC = be × EC

Certification:
∵AB∥DC,∠B＝90°
∴∠C＝∠B＝90
∴∠BAE+∠ABE＝90
∵∠AED＝90
∴∠DEC+∠AEB＝180-∠AED＝90
∴∠DEC＝∠BAE
∴△ABE∽△ECD
∴AB/BE＝EC/DC
∴AB×DC＝BE×EC

Known: as shown in the figure, in diamond ABCD, point E is on diagonal AC, point F is on the extension line of BC, EF = EB, EF and CD intersect at point g. (1) prove: eg · GF = CG · GD; (2) connect DF, if EF ⊥ CD, then what is the quantitative relationship between ∠ FDC and ∠ ADC? Prove your conclusion

(1) It is proved that: connecting ed, (1 point) ∵ point E is on the diagonal AC of diamond ABCD, with ∵ ECB = ∵ ECD, (2 points) ∵ BC = CD, CE = CE, ≌ BCE ≌ DCE; (3 points) ∵ EDC = ≌ EBC, (4 points) ∵ EB = EF, ∵ EBC = ∩ EFC; (5 points) ∵ EDC = ≌ EFC; (6 points) ∵ DGE =