證明：如圖，連接$OD$、$BD。$
$\because AB$為$\odot O$的直徑，
$\therefore \angle ADB = 90^{\circ}，$即$BD\perp AC。$
$\because AB = CB，$
$\therefore AD = CD。$
$\because OA = OB，$
$\therefore OD$為$\triangle ABC$的中位線，
$\therefore OD// BC，$
$\therefore \angle ODE=\angle DEC。$
$\because DE\perp BC，$
$\therefore \angle DEC = 90^{\circ}，$
$\therefore \angle ODE = 90^{\circ}，$即$DF\perp OD。$
$\because OD$為$\odot O$的半徑，
$\therefore DF$為$\odot O$的切線。
