证明:(Ⅰ)设以A(x1,x12)为切点的切线方程为y﹣x12=k(x﹣x1),
联立抛物线方程,可得x2﹣kx+kx1﹣x12=0,
由△=k2﹣4kx1+4x12=(k
orange key ring is Timmy's. It's present from Neil.
如图,△ABC沿直线l向右移了3厘米,得△FDE,且BC=6厘米,∠B=40°.
(1)求BE;
(2)求∠FDB的度数;
(3)找出图中相等的线段(不另添加线段);
(4)找出图中互相平行的线段(不另添加线段)