As shown in the figure, in isosceles trapezoid ABCD, ad ‖ BC. AB = DC, e is the midpoint of BC, connecting AE and De, proving: AE = De

As shown in the figure, in isosceles trapezoid ABCD, ad ‖ BC. AB = DC, e is the midpoint of BC, connecting AE and De, proving: AE = De


It is proved that: ∵ quadrilateral ABCD is isosceles trapezoid, ∵ AB = DC, ∵ B = ∵ C. ∵ e is the midpoint of BC, ∵ be = CE. In △ Abe and △ DCE, ab = DC, ≌ B = ≌ CBE = CE, ≌ Abe ≌ DCE (SAS) ≌ AE = De



As shown in the figure, in the isosceles trapezoid ABCD, ad ∥ BC, point E is the midpoint of BC side


It is proved that: ∵ quadrilateral ABCD is isosceles trapezoid, ∵ AB = DC, ∵ B = ∵ C. ∵ e is the midpoint of BC, ∵ be = CE. In △ Abe and △ DCE, ab = DC, ≌ B = ≌ CBE = CE, ≌ Abe ≌ DCE (SAS), ≌ AE = De



It is known that: as shown in the figure, the diagonal of rectangle ABCD intersects at O, AE bisects ∠ bad intersects BC at e, ∠ CAE = 15 °, then ∠ BOE=______ °.


∵ AE bisection ∠ bad intersects BC with E, ∵ AEB = 45 °, ab = be, ∵ CAE = 15 °, ∵ ACB = ∠ AEB - ∠ CAE = 45 ° - 15 ° = 30 °, ∵ Bao = 60 ° and ∵ OA = ob, ∵ boa is equilateral triangle, ∵ OA = OB = AB, that is ob = AB = be, ∵ BOE is isosceles triangle, and ∵ OBE = ∠ OCB = 30



As shown in the figure: given rectangular ABCD, AC and BD intersect at O, AE ‖ BD, de ‖ AC. verification: OE ⊥ ad


It is proved that: ∵ AC, BD are diagonals of rectangles, ∵ AC = BD, Ao = do, ∵ AE ∥ BD, de ∥ AC, ∵ quadrilateral Aode is parallelogram, ∵ Ao = do, ∵ quadrilateral Aode is diamond, ∵ OE ⊥ ad