⑴证明：∵CE是中线,E是AB中点,
∵AD⊥BC,∴DE=1/2AB=BE,
∴CD=BE=DE,
∴ΔCDE是等腰三角形.
⑵∵DC=DE,∴∠DCE=∠DEC,
∴∠BDE=∠DEC+∠DCE=2∠DCE,
∵DE=BE,
∴∠B=∠BDE=2∠BCE.