1. As shown in the figure, e is a point in the square ABCD, and △ Abe is an equilateral triangle. Think about the relationship between ∠ DCE and ∠ CEB, and explain the reason. 2 1. As shown in the figure, e is a point in the square ABCD, and △ Abe is an equilateral triangle. Think about the relationship between ∠ DCE and ∠ CEB, and explain the reason 2. As shown in the figure, the upper bottom ad of the isosceles trapezoid ABCD = 1, the lower bottom BC = 3, the diagonal AC ⊥ BD, find the height and length of the isosceles trapezoid

1. As shown in the figure, e is a point in the square ABCD, and △ Abe is an equilateral triangle. Think about the relationship between ∠ DCE and ∠ CEB, and explain the reason. 2 1. As shown in the figure, e is a point in the square ABCD, and △ Abe is an equilateral triangle. Think about the relationship between ∠ DCE and ∠ CEB, and explain the reason 2. As shown in the figure, the upper bottom ad of the isosceles trapezoid ABCD = 1, the lower bottom BC = 3, the diagonal AC ⊥ BD, find the height and length of the isosceles trapezoid

1. There is mutual complementary between ∠ DCE and ∠ CEB, that is, DCE + CEB = 90 & # 186
(∵ ABCD is a square, △ Abe is an equilateral triangle, ∵ BC = AB = be, so ∵ BCE = ∵ Dec.)
2. Let AC ⊥ BD be equal to e, then be = CE is known from ABCD as isosceles trapezoid, thus ∠ BCE = ∠ CBE = 45 & # 186;,
If EF ⊥ BC is set at F through E and ad is set at g through e, it is easy to know that f and G are the midpoint of BC and ad respectively,
∵AD//BC,∴∠ADE=∠CBE=45º,
The height of this isosceles trapezoid is FG = Fe + eg = FC / 2 + GD / 2 = (BC + AD) / 2 = 2
Let DH ⊥ BC be in H, then DH = FG = 2, FH = GD = 1 / 2, ch = fc-fh = 1,
In RT △ DHC, the Pythagorean theorem gives DC = √ 5. That is to say, the waist length of this isosceles trapezoid is √ 5