證明:延長(zhǎng)$AD$至點(diǎn)$E,$使$DE = AD,$連接$BE。$
$\because AD$是$\triangle ABC$的中線,
$\therefore BD = CD。$
在$\triangle BDE$和$\triangle CDA$中,
$\begin{cases} BD = CD \\ \angle BDE = \angle CDA \\ DE = DA \end{cases}$
$\therefore \triangle BDE \cong \triangle CDA(SAS)。$
$\therefore BE = AC。$
在$\triangle ABE$中,$AB + BE > AE。$
$\because AE = AD + DE = 2AD,$$BE = AC,$
$\therefore AB + AC > 2AD。$