典例1（2023·宿迁泗洪二模）如图，△ABC≌△DEF，CD平分∠BCA. 若∠A = 28°，∠CGF = 88°，则∠E的度数是（ ）

A. 32°
B. 34°
C. 40°
D. 44°

$\because \triangle ABC\cong \triangle DEF$，$\therefore \angle A=\angle D$，$\angle B=\angle E$．
$\because \angle A = 28^{\circ}$，$\therefore \angle D = 28^{\circ}$．$\because \angle CGF=\angle D+\angle DCB = 88^{\circ}$，
$\therefore \angle DCB=\angle CGF-\angle D=88^{\circ}-28^{\circ}=60^{\circ}$．$\because CD$平分$\angle BCA$，$\therefore \angle BCA = 2\angle DCB = 120^{\circ}$．$\therefore \angle E=\angle B = 180^{\circ}-\angle A-\angle BCA=180^{\circ}-28^{\circ}-120^{\circ}=32^{\circ}$．故选A．

【变式】如图，△ABC≌△ADE，∠B = 70°，∠C = 30°，∠DAC = 20°，则∠EAC的度数为_______.

$60^{\circ}$．

典例2（2023·宿迁宿豫三模）如图，AB = AC，添加一个条件，不能使△ABF≌△ACE的是（ ）

A. AE = AF
B. ∠B = ∠C
C. ∠AEC = ∠AFB
D. CE = BF

D．

典例3（2023·淮安）如图，D为线段CB上一点，BD = AC，∠E = ∠ABC，DE//AC. 求证：DE = CB.

$\because DE// AC$，$\therefore \angle EDB=\angle C$．在$\triangle BDE$和$\triangle ACB$中，$\begin{cases} \angle E=\angle ABC, \\ \angle EDB=\angle C, \\ BD = AC, \end{cases}$ $\therefore \triangle BDE\cong \triangle ACB(AAS)$．$\therefore DE = CB$．

典例4（2024·苏州昆山一模）如图，在△ABC中，D为边AB上一点，E为边AC的中点，连接DE并延长至点F，使得EF = ED，连接BE，CF.
（1）求证：CF//AB；
（2）若∠A = 70°，∠F = 35°，BE⊥AC，求∠BED的度数.

(1) $\because E$为边$AC$的中点，$\therefore AE = CE$．在$\triangle AED$和$\triangle CEF$中，$\begin{cases} AE = CE, \\ \angle AED=\angle CEF, \\ DE = FE, \end{cases}$ $\therefore \triangle AED\cong \triangle CEF(SAS)$．$\therefore \angle A=\angle ECF$．$\therefore CF// AB$．(2) 由(1)知，$\angle A=\angle ECF = 70^{\circ}$．又$\because \angle F = 35^{\circ}$，$\therefore \angle AED=\angle CEF = 180^{\circ}-\angle ECF-\angle F=180^{\circ}-70^{\circ}-35^{\circ}=75^{\circ}$．$\because BE\perp AC$，$\therefore \angle AEB = 90^{\circ}$．$\therefore \angle BED=\angle AEB-\angle AED=90^{\circ}-75^{\circ}=15^{\circ}$．