1. AD 是$\triangle ABC$的中线,$AB = 8,AC = 4$,则 AD 的取值范围是______.

$2 ＜ AD ＜ 6$

2. 如图,$OA = OB,OC = OD,∠AOB = ∠COD = 90^{\circ }$,取 BD 的中点 P,连接 OP. 求证:$AC = 2OP$.

证明:延长 $OP$ 至点 $E$,使 $PE = OP$,连接 $BE$.
$\because P$ 为 $BD$ 的中点,
$\therefore BP = PD$,
$\therefore \triangle BPE \cong \triangle DPO(SAS)$,
$\therefore BE = OD$, $\angle E = \angle DOP$,
$\therefore BE // OD$,
$\therefore \angle EBO + \angle BOD = 180^{\circ}$.
$\because \angle AOB = \angle COD = 90^{\circ}$,
$\therefore \angle BOD + \angle AOC = 360^{\circ} - 180^{\circ} = 180^{\circ}$,
$\therefore \angle EBO = \angle AOC$.
$\because BE = OD$, $OD = OC$,
$\therefore BE = OC$,
$\therefore \triangle EBO \cong \triangle COA(SAS)$,
$\therefore OE = AC$.
又 $\because OE = 2OP$,
$\therefore AC = 2OP$.
![img alt=3]

3. 如图,AD 是$\triangle ABC$的中线,E,F 分别在 AB,AC 上,且$DE⊥DF$. 求证:$BE + CF＞EF$.

证明:延长 $FD$ 至点 $G$,使 $GD = DF$,连接 $BG$, $EG$.
$\because DE \perp DF$,
$\therefore \angle EDF = \angle EDG = 90^{\circ}$,
$\therefore \triangle EDF \cong \triangle EDG(SAS)$,
$\therefore EF = EG$.
$\because BD = CD$, $\angle BDG = \angle CDF$, $GD = DF$,
$\therefore \triangle DFC \cong \triangle DGB(SAS)$,
$\therefore BG = CF$.
在 $\triangle BEG$ 中, $BG + BE ＞ EG$.
又 $\because EF = EG$, $BG = CF$,
$\therefore BE + CF ＞ EF$.

4. 如图,$AD = AE,AB = AC$,F 为 BE 的中点,$∠EAF = ∠ADC = 90^{\circ }$. 求证:$CD = 2AF$.

证明:过点 $B$ 作 $AF$ 的垂线,交 $AF$ 的延长线于点 $M$.
则 $\angle EAF = \angle M = 90^{\circ}$.
在 $\triangle EAF$ 与 $\triangle BMF$ 中,
$\angle EAF = \angle M$,
$\angle EFA = \angle MFB$, $EF = BF$,
$\therefore \triangle EAF \cong \triangle BMF(AAS)$,
$\therefore AF = MF$, $BM = AE = AD$.
在 $Rt\triangle ADC$ 与 $Rt\triangle BMA$ 中,
$AB = AC$, $BM = AD$,
$\therefore Rt\triangle ADC \cong Rt\triangle BMA(HL)$,
$\therefore DC = AM = 2AF$.