证明：如图，连接$OD$、$BD。$
$\because AB$为$\odot O$的直径，
$\therefore \angle ADB = 90^{\circ}，$即$BD\perp AC。$
$\because AB = CB，$
$\therefore AD = CD。$
$\because OA = OB，$
$\therefore OD$为$\triangle ABC$的中位线，
$\therefore OD// BC，$
$\therefore \angle ODE=\angle DEC。$
$\because DE\perp BC，$
$\therefore \angle DEC = 90^{\circ}，$
$\therefore \angle ODE = 90^{\circ}，$即$DF\perp OD。$
$\because OD$为$\odot O$的半径，
$\therefore DF$为$\odot O$的切线。
