【例2】如图,在等边三角形ABC中,$DE// BC$,求证：$\triangle ADE$是等边三角形。
 

证明：

【解析】：根据等边三角形性质得出$\angle A = \angle B = \angle C = 60^{\circ}$，再由平行线性质得到$\angle ADE = \angle B = 60^{\circ}$，$\angle AED = \angle C = 60^{\circ}$，从而得出$\angle A = \angle ADE = \angle AED = 60^{\circ}$，最后根据等边三角形判定定理得出$\triangle ADE$是等边三角形。
【答案】：证明：$\because \triangle ABC$是等边三角形，$\therefore \angle A = \angle B = \angle C = 60^{\circ}$。$\because DE// BC$，$\therefore \angle ADE = \angle B = 60^{\circ}$，$\angle AED = \angle C = 60^{\circ}$，$\therefore \angle A = \angle ADE = \angle AED = 60^{\circ}$，$\therefore \triangle ADE$是等边三角形。

【练2】如图,$\triangle ABC$是等边三角形,点P在$\triangle ABC$的内部,点Q在$\triangle ABC$的外部,且$∠ABP= ∠ACQ,BP= CQ$,连接AP,AQ。问：$\triangle APQ$是什么形状的三角形？试说明你的理由。

解：$\triangle APQ$是.
理由：$\because \triangle ABC$是等边三角形，
$\therefore AB=AC,\angle BAC=60^{\circ }.$
在$\triangle ABP$和$\triangle ACQ$中，$\left\{\begin{array}{l} AB=AC,\\ \angle ABP=\angle ACQ,\\ BP=CQ,\end{array}\right. $
$\therefore \triangle ABP\cong \triangle ACQ$,
$\therefore AP=AQ,\angle BAP=\angle CAQ.$
$\because \angle BAC=\angle BAP+\angle PAC=60^{\circ },$
$\therefore \angle PAQ=\angle CAQ+\angle PAC=60^{\circ },$
$\therefore \triangle APQ$是等边三角形.

解：$\triangle APQ$是等边三角形.
理由：$\because \triangle ABC$是等边三角形，
$\therefore AB=AC,\angle BAC=60^{\circ }.$
在$\triangle ABP$和$\triangle ACQ$中，$\left\{\begin{array}{l} AB=AC,\\ \angle ABP=\angle ACQ,\\ BP=CQ,\end{array}\right. $
$\therefore \triangle ABP\cong \triangle ACQ(SAS),$
$\therefore AP=AQ,\angle BAP=\angle CAQ.$
$\because \angle BAC=\angle BAP+\angle PAC=60^{\circ },$
$\therefore \angle PAQ=\angle CAQ+\angle PAC=60^{\circ },$
$\therefore \triangle APQ$是等边三角形.