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第57页

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$82^{\circ}$

①②③⑤

 （1）证明：

 $\because BF = AC，$$AB = AE，$

 $\therefore BF + AB = AC + AE，$即$FA = EC.$

 $\because \triangle DEF$是等边三角形，

 $\therefore EF = DE.$

 在$\triangle AEF$和$\triangle CDE$中，

 $\begin{cases}AE = CD\\FA = EC\\EF = DE\end{cases}，$

 $\therefore \triangle AEF\cong\triangle CDE(SSS).$

 （2）证明：

 由（1），得$\triangle AEF\cong\triangle CDE，$

 $\therefore \angle FEA=\angle EDC.$

 $\because \angle BCA=\angle EDC+\angle DEC，$$\triangle DEF$是等边三角形，

 $\therefore \angle BCA=\angle FEA+\angle DEC=\angle DEF = 60^{\circ}.$

 同理，可得$\angle BAC = 60^{\circ}.$

 $\therefore \angle ABC=\angle BAC=\angle BCA = 60^{\circ}.$

 $\therefore \triangle ABC$是等边三角形.

 （1）解：

 $\because AP = AQ，$

 $\therefore \angle APQ=\angle AQP.$

 $\therefore \angle APB=\angle AQC.$

 在$\triangle APB$和$\triangle AQC$中，

 $\begin{cases}AP = AQ\\\angle APB=\angle AQC\\BP = CQ\end{cases}，$

 $\therefore \triangle APB\cong\triangle AQC(SAS).$

 $\therefore AB = AC.$

 $\therefore \angle B=\angle C = 25^{\circ}.$

 $\therefore \angle BAC=180^{\circ}-(\angle B+\angle C)=130^{\circ}.$

 $\because AP = BP，$$AQ = CQ，$

 $\therefore \angle BAP=\angle B = 25^{\circ}，$$\angle CAQ=\angle C = 25^{\circ}.$

 $\therefore \angle PAQ=\angle BAC-\angle BAP-\angle CAQ = 80^{\circ}.$

 （2）解：

 由（1），知$\angle B=\angle C.$

 $\because \angle BAC = 120^{\circ}，$

 $\therefore \angle B=\angle C = 30^{\circ}.$

 同（1），得$\angle BAP=\angle CAQ=\angle B=\angle C = 30^{\circ}，$

 $\therefore$易得$\angle PAQ = 60^{\circ}.$

 又$\because AP = AQ，$

 $\therefore \triangle APQ$是等边三角形.

 （1）证明：

 如图①，在$AC$上截取$CM = CD，$连接$DM.$

 $\because \triangle ABC$是等边三角形，

 $\therefore \angle ACB = 60^{\circ}.$

 $\therefore \triangle CDM$是等边三角形.

 $\therefore MD = CD = CM，$$\angle CMD=\angle CDM = 60^{\circ}.$

 $\therefore \angle AMD=180^{\circ}-\angle CMD = 120^{\circ}.$

 $\because \angle ADE = 60^{\circ}，$

 $\therefore \angle ADE=\angle MDC.$

 $\therefore \angle ADE-\angle MDE=\angle MDC-\angle MDE，$即$\angle ADM=\angle EDC.$

 $\because DE$与$\angle ACB$的邻补角的平分线交于点$E，$

 $\therefore$易得$\angle ACE = 60^{\circ}.$

 $\therefore \angle ECD=\angle ACE+\angle ACB = 120^{\circ}=\angle AMD.$

 在$\triangle ADM$和$\triangle EDC$中，

 $\begin{cases}\angle ADM=\angle EDC\\MD = CD\\\angle AMD=\angle ECD\end{cases}，$

 $\therefore \triangle ADM\cong\triangle EDC(ASA).$

 $\therefore AM = EC.$

 $\therefore CA = CM + AM = CD + CE.$

 （2）解：$CA = CE - CD.$

 证明：如图②，延长$AC，$在$AC$的延长线上截取$CM = CD，$连接$DM.$

 $\because \triangle ABC$是等边三角形，

 $\therefore \angle ACB = 60^{\circ}.$

 $\therefore \angle DCM=\angle ACB = 60^{\circ}.$

 $\therefore \triangle CDM$是等边三角形.

 $\therefore MD = CD = CM，$$\angle M=\angle CDM = 60^{\circ}.$

 $\because DE$与$\angle ACB$的邻补角的平分线交于点$E，$

 $\therefore$易得$\angle ACE=\angle ECD = 60^{\circ}.$

 $\therefore \angle ECD=\angle M.$

 $\because \angle ADE = 60^{\circ}，$

 $\therefore \angle ADE=\angle CDM.$

 $\therefore \angle CDM+\angle ADC=\angle ADE+\angle ADC，$即$\angle ADM=\angle EDC.$

 在$\triangle ADM$和$\triangle EDC$中，

 $\begin{cases}\angle ADM=\angle EDC\\MD = CD\\\angle M=\angle ECD\end{cases}，$

 $\therefore \triangle ADM\cong\triangle EDC(ASA).$

 $\therefore AM = EC.$

 $\therefore CA = AM - CM = CE - CD.$