2. 如图,在 $\mathrm{Rt}△ ABC$ 中,$∠ BAC=90°$,$∠ C=30°$,$AD⊥ BC$ 于点 $D$,$BF$ 平分 $∠ ABC$,交 $AD$ 于点 $E$,交 $AC$ 于点 $F$.
(1)求证:$△ AEF$ 是等边三角形.
(2)若 $BC=4$,则 $BD$ 的长为 ______.

1
; ​$ (1) $​证明：∵​$∠BAC = 90°，$​​$∠C = 30°，$​∴​$∠ABC = 180° - ∠BAC - ∠C = 180° - 90° - 30° = 60°.$​∵​$BF $​平分​$∠ABC，$​∴​$∠ABF = ∠CBF = \frac {1}{2}∠ABC = \frac {1}{2}×60° = 30°.$​∵​$AD⊥BC，$​∴​$∠ADB = 90°，$​∴​$∠AEF = ∠BED = 90° - ∠CBF = 90° - 30° = 60°.$​又∵​$∠AFB = 90° - ∠ABF = 90° - 30° = 60°，$​∴​$∠AEF = ∠AFB，$​∴​$△AEF $​是等边三角形​$.$​

3. 如图,在$△ ABC$中,$∠ A = ∠ BCA = 15°$,$AB = 5$,求$△ ABC$的面积.

解：如图，过点$C$作$CD\perp AB$交$AB$的延长线于点$D。$

$\because\angle A = \angle BCA = 15^{\circ}，$$AB = 5，$
$\therefore BC = AB = 5，$
$\angle DBC = \angle A + \angle BCA = 15^{\circ}+15^{\circ}=30^{\circ}，$
$\therefore CD = \frac{1}{2}BC = \frac{5}{2}，$
$\therefore S_{\triangle ABC}=\frac{1}{2}AB\cdot CD=\frac{1}{2}\times5\times\frac{5}{2}=\frac{25}{4}。$

4. 如图,在$△ ABC$中,$M$、$N$分别是$BC$与$EF$的中点,$CF ⊥ AB$,$BE ⊥ AC$.
(1)求证:$MN ⊥ EF$.
(2)已知$BC=8$,当$∠ A=60°$时,求$EF$的长.

(1)证明：如图，连接$FM$、$EM。$

$\because CF\perp AB，$$BE\perp AC，$
$\therefore\angle CEB = \angle CFB = 90^{\circ}。$
$\because M$是$BC$的中点，$\therefore FM = BM = CM = \frac{1}{2}BC = EM。$
又$\because N$是$EF$的中点，$\therefore MN\perp EF。$ (2)$\because\angle A = 60^{\circ}，$
$\therefore\angle ABC + \angle ACB = 180^{\circ}-\angle A = 180^{\circ}-60^{\circ}=120^{\circ}。$
由 (1)，得$BM = FM = CM = EM = \frac{1}{2}BC = 4，$
$\therefore\angle ABC = \angle BFM，$$\angle ACB = \angle CEM，$$\therefore\angle BFM + \angle CEM = 120^{\circ}，$$\therefore\angle FMB + \angle EMC = 360^{\circ}-(\angle ABC + \angle ACB+\angle BFM + \angle CEM)=360^{\circ}-(120^{\circ}+120^{\circ}) = 120^{\circ}，$
$\therefore\angle EMF = 180^{\circ}-(\angle FMB + \angle EMC)=180^{\circ}-120^{\circ}=60^{\circ}。$
$\because FM = EM = 4，$
$\therefore\triangle EMF$为等边三角形，
$\therefore EF = FM = 4。$

1. 如图,在 $Rt△ ABC$ 中,$∠ C=90°$.
(1)求作$∠ BAC$ 的平分线,与 $BC$ 交于点 $D$.(用尺规作图,不写作法,保留作图痕迹)
(2)若 $CD=4$,$AB=16$,求$△ ABD$ 的面积.

(1)如图，$AD$即为所求。
(2)如图，过点$D$作$DE\perp AB$于点$E。$
$\because AD$平分$\angle BAC，$$\angle C = 90^{\circ}，$
$\therefore DE = DC = 4，$
$\therefore S_{\triangle ABD}=\frac{1}{2}AB\cdot DE=\frac{1}{2}\times16\times4 = 32。$