23. (8分)如图,已知$\triangle ABC$,$\angle ABC的平分线与\triangle ABC的外角\angle ACN的平分线相交于点P$,连接$AP$.
(1) 求证:$AP平分\angle CAM$;
(2) 过点$C作CE\perp AP$,$E$是垂足,并延长$CE交BM于点D$.求证:$CE = ED$.
(1) 求证:$AP平分\angle CAM$;
(2) 过点$C作CE\perp AP$,$E$是垂足,并延长$CE交BM于点D$.求证:$CE = ED$.
答案：
(1)证明:过点P作$PT\perp BN$于点T,$PS\perp AC$于点S,$PQ\perp BM$于点Q,如答图.
∵$\angle ABC$的平分线与$\triangle ABC$的外角$\angle ACN$的平分线相交于点P,
∴$PQ = PT$,$PS = PT$,∴$PQ = PS$.
∴AP平分$\angle CAM$.
(2)∵AP平分$\angle CAM$,∴$\angle DAE=\angle CAE$.
∵$CE\perp AP$,∴$\angle AED=\angle AEC = 90^{\circ}$.
在$\triangle AED$和$\triangle AEC$中,$\begin{cases}\angle DAE = \angle CAE\\AE = AE\\\angle DEA = \angle CEA\end{cases}$
∴$\triangle AED\cong\triangle AEC$,
∴$CE = ED$.
(1)证明:过点P作$PT\perp BN$于点T,$PS\perp AC$于点S,$PQ\perp BM$于点Q,如答图.
∵$\angle ABC$的平分线与$\triangle ABC$的外角$\angle ACN$的平分线相交于点P,
∴$PQ = PT$,$PS = PT$,∴$PQ = PS$.
∴AP平分$\angle CAM$.
(2)∵AP平分$\angle CAM$,∴$\angle DAE=\angle CAE$.
∵$CE\perp AP$,∴$\angle AED=\angle AEC = 90^{\circ}$.
在$\triangle AED$和$\triangle AEC$中,$\begin{cases}\angle DAE = \angle CAE\\AE = AE\\\angle DEA = \angle CEA\end{cases}$
∴$\triangle AED\cong\triangle AEC$,
∴$CE = ED$.
24. (10分)如图,$\triangle ABC$为等边三角形,$P为BC$上一点,$\triangle APQ$为等边三角形.
(1) 求证:$AB// CQ$;
(2) 是否存在点$P使得AQ\perp CQ$? 若存在,指出点$P$的位置;若不存在,请说明理由.
(1) 求证:$AB// CQ$;
(2) 是否存在点$P使得AQ\perp CQ$? 若存在,指出点$P$的位置;若不存在,请说明理由.
答案：(1)证明:∵$\triangle ABC$和$\triangle APQ$都是等边三角形,
∴$AB = AC$,$AP = AQ$,$\angle B=\angle BAC=\angle PAQ = 60^{\circ}$.
∴$\angle BAC-\angle PAC=\angle PAQ-\angle PAC$,
∴$\angle BAP=\angle CAQ$.
在$\triangle ABP$和$\triangle ACQ$中,$\begin{cases}AB = AC\\\angle BAP = \angle CAQ\\AP = AQ\end{cases}$
∴$\triangle ABP\cong\triangle ACQ(SAS)$.
∴$\angle ACQ=\angle B=\angle BAC = 60^{\circ}$,∴$AB// CQ$.
(2)解:存在，当P为BC的中点时,$AQ\perp CQ$.
由(1)知,$\triangle ABP\cong\triangle ACQ$,
∴$\angle APB=\angle AQC$.
∵P为BC的中点，$\triangle ABC$为等边三角形,
∴$AP\perp BC$,即$\angle APB = 90^{\circ}$,
∴$\angle AQC = 90^{\circ}$,
∴$AQ\perp QC$.
即当P为BC的中点时,$AQ\perp CQ$.
∴$AB = AC$,$AP = AQ$,$\angle B=\angle BAC=\angle PAQ = 60^{\circ}$.
∴$\angle BAC-\angle PAC=\angle PAQ-\angle PAC$,
∴$\angle BAP=\angle CAQ$.
在$\triangle ABP$和$\triangle ACQ$中,$\begin{cases}AB = AC\\\angle BAP = \angle CAQ\\AP = AQ\end{cases}$
∴$\triangle ABP\cong\triangle ACQ(SAS)$.
∴$\angle ACQ=\angle B=\angle BAC = 60^{\circ}$,∴$AB// CQ$.
(2)解:存在，当P为BC的中点时,$AQ\perp CQ$.
由(1)知,$\triangle ABP\cong\triangle ACQ$,
∴$\angle APB=\angle AQC$.
∵P为BC的中点，$\triangle ABC$为等边三角形,
∴$AP\perp BC$,即$\angle APB = 90^{\circ}$,
∴$\angle AQC = 90^{\circ}$,
∴$AQ\perp QC$.
即当P为BC的中点时,$AQ\perp CQ$.
25. (10分)(2023春·高新区期末)如图,在$\triangle ABC$中,$AD为BC$边上的高,$AE是\angle BAD$的平分线,点$F为AE$上一点,连接$BF$,$\angle BFE = 45^{\circ}$.
(1) 求证:$BF平分\angle ABE$;
(2) 连接$CF交AD于点G$,若$S_{\triangle ABF} = S_{\triangle CBF}$,求证:$\angle AFC = 90^{\circ}$;
(3) 在(2)的条件下,当$BE = 3$,$AG = 4.5$时,求线段$AB$的长.
(1) 求证:$BF平分\angle ABE$;
(2) 连接$CF交AD于点G$,若$S_{\triangle ABF} = S_{\triangle CBF}$,求证:$\angle AFC = 90^{\circ}$;
(3) 在(2)的条件下,当$BE = 3$,$AG = 4.5$时,求线段$AB$的长.
答案：
(1)证明:∵AE是$\angle BAD$的平分线,
∴$\angle BAD = 2\angle BAF$.
∵$\angle BFE = 45^{\circ}$,
∴$\angle FBA+\angle BAF = 45^{\circ}$,∴$2\angle FBA+2\angle BAF = 90^{\circ}$.
∵AD为BC边上的高,
∴$\angle EBF+\angle FBA+2\angle BAF = 90^{\circ}$,
∴$2\angle FBA=\angle EBF+\angle FBA$,∴$\angle EBF=\angle FBA$.
∴BF平分$\angle ABE$.
(2)证明:如答图,过点F作$FM\perp BC$于点M,$FN\perp AB$于点N.
∵BF平分$\angle ABE$,$FM\perp BC$,$FN\perp AB$,∴$FM = FN$.
∵$S_{\triangle ABF}=S_{\triangle CBF}$,即$\frac{1}{2}AB\cdot FN=\frac{1}{2}BC\cdot FM$,
∴$AB = BC$.
在$\triangle ABF$和$\triangle CBF$中,$\begin{cases}BA = BC\\\angle ABF = \angle CBF\\BF = BF\end{cases}$
∴$\triangle ABF\cong\triangle CBF(SAS)$.
∴$\angle AFB=\angle CFB$.
∵$\angle BFE = 45^{\circ}$,∴$\angle AFB = 135^{\circ}$,
∴$\angle CFB = 135^{\circ}$,∴$\angle CFE=\angle CFB-\angle BFE = 135^{\circ}-45^{\circ}=90^{\circ}$,∴$\angle AFC = 90^{\circ}$.
(3)解:∵$\triangle ABF\cong\triangle CBF$,∴$AF = FC$,$AB = BC$.
∵$\angle AFC=\angle ADC = 90^{\circ}$,$\angle AGF=\angle CGD$,
∴$\angle FAG=\angle FCE$.
在$\triangle AFG$和$\triangle CFE$中,$\begin{cases}\angle AFG = \angle CFE\\AF = CF\\\angle FAG = \angle FCE\end{cases}$
∴$\triangle AFG\cong\triangle CFE(ASA)$.
∴$AG = EC = 4.5$.
∵$BE = 3$,∴$BC = BE + EC = 7.5$.
∴$AB = BC = 7.5$.
(1)证明:∵AE是$\angle BAD$的平分线,
∴$\angle BAD = 2\angle BAF$.
∵$\angle BFE = 45^{\circ}$,
∴$\angle FBA+\angle BAF = 45^{\circ}$,∴$2\angle FBA+2\angle BAF = 90^{\circ}$.
∵AD为BC边上的高,
∴$\angle EBF+\angle FBA+2\angle BAF = 90^{\circ}$,
∴$2\angle FBA=\angle EBF+\angle FBA$,∴$\angle EBF=\angle FBA$.
∴BF平分$\angle ABE$.
(2)证明:如答图,过点F作$FM\perp BC$于点M,$FN\perp AB$于点N.
∵BF平分$\angle ABE$,$FM\perp BC$,$FN\perp AB$,∴$FM = FN$.
∵$S_{\triangle ABF}=S_{\triangle CBF}$,即$\frac{1}{2}AB\cdot FN=\frac{1}{2}BC\cdot FM$,
∴$AB = BC$.
在$\triangle ABF$和$\triangle CBF$中,$\begin{cases}BA = BC\\\angle ABF = \angle CBF\\BF = BF\end{cases}$
∴$\triangle ABF\cong\triangle CBF(SAS)$.
∴$\angle AFB=\angle CFB$.
∵$\angle BFE = 45^{\circ}$,∴$\angle AFB = 135^{\circ}$,
∴$\angle CFB = 135^{\circ}$,∴$\angle CFE=\angle CFB-\angle BFE = 135^{\circ}-45^{\circ}=90^{\circ}$,∴$\angle AFC = 90^{\circ}$.
(3)解:∵$\triangle ABF\cong\triangle CBF$,∴$AF = FC$,$AB = BC$.
∵$\angle AFC=\angle ADC = 90^{\circ}$,$\angle AGF=\angle CGD$,
∴$\angle FAG=\angle FCE$.
在$\triangle AFG$和$\triangle CFE$中,$\begin{cases}\angle AFG = \angle CFE\\AF = CF\\\angle FAG = \angle FCE\end{cases}$
∴$\triangle AFG\cong\triangle CFE(ASA)$.
∴$AG = EC = 4.5$.
∵$BE = 3$,∴$BC = BE + EC = 7.5$.
∴$AB = BC = 7.5$.