1.(教材$P_{63}T_{11}$变式)如图,$\triangle ABC$为等边三角形,$D为\triangle ABC$外的一点,且$∠ADC= 30^{\circ }$,连接$BD$.
(1)画出将$\triangle BAD绕点A逆时针旋转60^{\circ }得到的\triangle CAE$;
(2)求证:$BD^{2}= CD^{2}+AD^{2}$.

解: (1) 如图所示;

(2) 连接 DE, 则$BD=CE$,
$\triangle ADE$为等边三角形,
$\therefore AD=DE,∠ADE=60^{\circ },$
又$∠ADC=30^{\circ },$
$\therefore ∠CDE=90^{\circ },$
$\therefore CD^{2}+DE^{2}=CE^{2},$
即$BD^{2}=CD^{2}+AD^{2}.$

2.(教材$P_{63}T_{11}$变式)如图,在$\triangle ABC$中,$∠ABC= 30^{\circ },\triangle ACD$是等边三角形.若$AB= 6,BC= 8$,求$BD$的长.

解: 将$\triangle ABD$绕点 A 顺时针旋转$60^{\circ }$得到$\triangle AEC$, 连接 BE,
$\therefore AB=AE,∠BAE=60^{\circ },$
$BD=CE,$
$\therefore \triangle ABE$是等边三角形,
$\therefore ∠EBC=60^{\circ }+30^{\circ }=90^{\circ },$
$\therefore EB⊥BC,$
$\therefore BD=CE=\sqrt {BE^{2}+BC^{2}}=10.$

3.如图,$P为等边\triangle ABC$内一点,连接$PA,PB,PC,∠APB= 150^{\circ },∠BPC= 90^{\circ }$.
(1)求$PA:PB:PC$的值;
(2)若$PA= 1$,求$\triangle ABC$的面积.

解: (1) 将$\triangle PAB$绕点 B 顺时针旋转$60^{\circ }$得到$\triangle DCB$, 连接 PD,
$\therefore ∠PBD=60^{\circ },PB=BD,$
$∠BDC=∠APB=150^{\circ },CD=PA,$
$\therefore \triangle PBD$是等边三角形,
$\therefore ∠PDB=∠BPD=60^{\circ },$
$PD=PB,$
$\therefore ∠PDC=∠BDC-∠BDP=90^{\circ }.$
$\because ∠BPC=90^{\circ },∠BPD=60^{\circ },$
$\therefore ∠DPC=30^{\circ }$, 在$Rt\triangle PCD$中,
$\therefore CD=\frac {1}{2}PC,PD=\sqrt {3}CD,$
$\therefore PA:PB:PC=1:\sqrt {3}:2;$
(2)$\because PA=1$, 由(1)得$PB=\sqrt {3},$
$PC=2,\because ∠BPC=90^{\circ },$
$\therefore BC=\sqrt {PB^{2}+PC^{2}}$
$=\sqrt {2^{2}+(\sqrt {3})^{2}}$
$=\sqrt {7},$
$\therefore S_{\triangle ABC}=\frac {\sqrt {3}}{4}BC^{2}=\frac {\sqrt {3}}{4}×7=\frac {7\sqrt {3}}{4}.$