上,点G在BA的延长线上,且CE=BK=AG.
⑴求证:①DE=DG;②DE⊥DG;
⑵尺规作图:以线段DE,DG为边作出正方形DEFG(要求:只保留作图痕迹,不写作法和证明);
⑶连接⑵中的KF,猜想并写出四边形CEFK是怎样的特殊四边形,并证明你的猜想;
⑷当




同类型试题

y = sin x, x∈R, y∈[–1,1],周期为2π,函数图像以 x = (π/2) + kπ 为对称轴
y = arcsin x, x∈[–1,1], y∈[–π/2,π/2]
sin x = 0 ←→ arcsin x = 0
sin x = 1/2 ←→ arcsin x = π/6
sin x = √2/2 ←→ arcsin x = π/4
sin x = 1 ←→ arcsin x = π/2


y = sin x, x∈R, y∈[–1,1],周期为2π,函数图像以 x = (π/2) + kπ 为对称轴
y = arcsin x, x∈[–1,1], y∈[–π/2,π/2]
sin x = 0 ←→ arcsin x = 0
sin x = 1/2 ←→ arcsin x = π/6
sin x = √2/2 ←→ arcsin x = π/4
sin x = 1 ←→ arcsin x = π/2

