19. (6分)如图,在$\triangle ABC$中,$AC ＞ AB$,射线$AD平分\angle BAC$,交$BC于点E$,点$F在边AB$的延长线上,$AF = AC$,连接$EF$.
(1) 求证:$\triangle AEC\cong\triangle AEF$;
(2) 若$\angle AEB = 50^{\circ}$,求$\angle BEF$的度数.
(1) 求证:$\triangle AEC\cong\triangle AEF$;
(2) 若$\angle AEB = 50^{\circ}$,求$\angle BEF$的度数.
答案：(1)证明:∵射线AD平分$\angle BAC$,∴$\angle CAE=\angle FAE$.
在$\triangle AEC$和$\triangle AEF$中,$\begin{cases}AC = AF\\\angle CAE = \angle FAE\\AE = AE\end{cases}$
∴$\triangle AEC\cong\triangle AEF(SAS)$.
(2)解:∵$\triangle AEC\cong\triangle AEF$,∴$\angle C=\angle F$.
∵$\angle AEB=\angle CAE+\angle C = 50^{\circ}$,∴$\angle FAE+\angle F = 50^{\circ}$.
∵$\angle FAE+\angle F+\angle AEB+\angle BEF = 180^{\circ}$,
∴$\angle BEF = 80^{\circ}$.
在$\triangle AEC$和$\triangle AEF$中,$\begin{cases}AC = AF\\\angle CAE = \angle FAE\\AE = AE\end{cases}$
∴$\triangle AEC\cong\triangle AEF(SAS)$.
(2)解:∵$\triangle AEC\cong\triangle AEF$,∴$\angle C=\angle F$.
∵$\angle AEB=\angle CAE+\angle C = 50^{\circ}$,∴$\angle FAE+\angle F = 50^{\circ}$.
∵$\angle FAE+\angle F+\angle AEB+\angle BEF = 180^{\circ}$,
∴$\angle BEF = 80^{\circ}$.
20. (6分)如图,$AD是\triangle ABC$的中线,$BE\perp AD$,垂足为$E$,$CF\perp AD$,交$AD的延长线于点F$,$G是DA$延长线上一点,连接$BG$.
(1) 求证:$BE = CF$;
(2) 若$BG = CA$,求证:$GA = 2DE$.
(1) 求证:$BE = CF$;
(2) 若$BG = CA$,求证:$GA = 2DE$.
答案：证明:(1)∵AD是$\triangle ABC$的中线，∴$BD = CD$.
∵$BE\perp AD$,$CF\perp AD$,∴$\angle BED=\angle CFD$.
在$\triangle BED$和$\triangle CFD$中,$\begin{cases}\angle BED = \angle CFD\\\angle BDE = \angle CDF\\BD = CD\end{cases}$
∴$\triangle BED\cong\triangle CFD(AAS)$,
∴$BE = CF$.
(2)在$Rt\triangle BGE$和$Rt\triangle CAF$中,$\begin{cases}BG = CA\\BE = CF\end{cases}$
∴$Rt\triangle BGE\cong Rt\triangle CAF(HL)$,
∴$GE = AF$,∴$AG = EF$.
∵$\triangle BED\cong\triangle CFD$,
∴$DE = DF$,∴$GA = 2DE$.
∵$BE\perp AD$,$CF\perp AD$,∴$\angle BED=\angle CFD$.
在$\triangle BED$和$\triangle CFD$中,$\begin{cases}\angle BED = \angle CFD\\\angle BDE = \angle CDF\\BD = CD\end{cases}$
∴$\triangle BED\cong\triangle CFD(AAS)$,
∴$BE = CF$.
(2)在$Rt\triangle BGE$和$Rt\triangle CAF$中,$\begin{cases}BG = CA\\BE = CF\end{cases}$
∴$Rt\triangle BGE\cong Rt\triangle CAF(HL)$,
∴$GE = AF$,∴$AG = EF$.
∵$\triangle BED\cong\triangle CFD$,
∴$DE = DF$,∴$GA = 2DE$.
21. (8分)如图,在$\triangle ABC$中,$\angle A = 40^{\circ}$,点$D$,$E分别在边AB$,$AC$上,$BD = BC = CE$,连接$CD$,$BE$.
(1) 若$\angle ABC = 80^{\circ}$,求$\angle BDC$,$\angle ABE$的度数;
(2) 写出$\angle BEC与\angle BDC$之间的关系,并说明理由.
(1) 若$\angle ABC = 80^{\circ}$,求$\angle BDC$,$\angle ABE$的度数;
(2) 写出$\angle BEC与\angle BDC$之间的关系,并说明理由.
答案：(1)解:∵$\angle ABC = 80^{\circ}$,$BD = BC$,
∴$\angle BDC=\angle BCD=\frac{1}{2}×(180^{\circ}-80^{\circ}) = 50^{\circ}$.
∵$\angle A+\angle ABC+\angle ACB = 180^{\circ}$,$\angle A = 40^{\circ}$,$\angle ABC = 80^{\circ}$,
∴$\angle ACB = 180^{\circ}-40^{\circ}-80^{\circ}=60^{\circ}$.
∵$CE = BC$,∴$\triangle BCE$是等边三角形,∴$\angle EBC = 60^{\circ}$,
∴$\angle ABE=\angle ABC-\angle EBC = 20^{\circ}$.
(2)$\angle BEC+\angle BDC = 110^{\circ}$.
理由:设$\angle BEC=\alpha$,$\angle BDC=\beta$.
则$\alpha=\angle A+\angle ABE = 40^{\circ}+\angle ABE$.
∵$CE = BC$,∴$\angle CBE=\angle BEC=\alpha$,
∴$\angle ABC=\angle ABE+\angle CBE=\angle A + 2\angle ABE = 40^{\circ}+2\angle ABE$.
在$\triangle BDC$中,$BD = BC$,
∴$\angle BDC+\angle BCD+\angle DBC = 2\beta+40^{\circ}+2\angle ABE = 180^{\circ}$,
∴$\beta = 70^{\circ}-\angle ABE$.
∴$\alpha+\beta = 40^{\circ}+\angle ABE+70^{\circ}-\angle ABE = 110^{\circ}$,
∴$\angle BEC+\angle BDC = 110^{\circ}$.
∴$\angle BDC=\angle BCD=\frac{1}{2}×(180^{\circ}-80^{\circ}) = 50^{\circ}$.
∵$\angle A+\angle ABC+\angle ACB = 180^{\circ}$,$\angle A = 40^{\circ}$,$\angle ABC = 80^{\circ}$,
∴$\angle ACB = 180^{\circ}-40^{\circ}-80^{\circ}=60^{\circ}$.
∵$CE = BC$,∴$\triangle BCE$是等边三角形,∴$\angle EBC = 60^{\circ}$,
∴$\angle ABE=\angle ABC-\angle EBC = 20^{\circ}$.
(2)$\angle BEC+\angle BDC = 110^{\circ}$.
理由:设$\angle BEC=\alpha$,$\angle BDC=\beta$.
则$\alpha=\angle A+\angle ABE = 40^{\circ}+\angle ABE$.
∵$CE = BC$,∴$\angle CBE=\angle BEC=\alpha$,
∴$\angle ABC=\angle ABE+\angle CBE=\angle A + 2\angle ABE = 40^{\circ}+2\angle ABE$.
在$\triangle BDC$中,$BD = BC$,
∴$\angle BDC+\angle BCD+\angle DBC = 2\beta+40^{\circ}+2\angle ABE = 180^{\circ}$,
∴$\beta = 70^{\circ}-\angle ABE$.
∴$\alpha+\beta = 40^{\circ}+\angle ABE+70^{\circ}-\angle ABE = 110^{\circ}$,
∴$\angle BEC+\angle BDC = 110^{\circ}$.
22. (8分)已知一个三角形的三条边的长分别为$n + 6$,$3n$,$n + 2$.($n$为正整数)
(1) 若这个三角形是等腰三角形,求它的三边长;
(2) 求出$n$的所有整数值.
(1) 若这个三角形是等腰三角形,求它的三边长;
(2) 求出$n$的所有整数值.
答案：(1)①如果$n + 2 = 3n$,解得$n = 1$,三角形三条边的长分别为3,3,7，不符合三角形三边关系.
②如果$n + 6 = 3n$,解得$n = 3$,三角形三条边的长分别为5,9,9,符合三角形三边关系.
综上所述，等腰三角形的三边长分别为5，9,9.
(2)由题意得$\begin{cases}3n+(n + 2)＞n + 6\\(n + 6)+(n + 2)＞3n\end{cases}$，解得$\frac{4}{3}＜n＜8$.
∵n为整数，
∴n的所有整数值是2,3,4,5,6,7.
②如果$n + 6 = 3n$,解得$n = 3$,三角形三条边的长分别为5,9,9,符合三角形三边关系.
综上所述，等腰三角形的三边长分别为5，9,9.
(2)由题意得$\begin{cases}3n+(n + 2)＞n + 6\\(n + 6)+(n + 2)＞3n\end{cases}$，解得$\frac{4}{3}＜n＜8$.
∵n为整数，
∴n的所有整数值是2,3,4,5,6,7.