1、(1)AD 95
(2)AC = A'C'或∠B = ∠B'或∠C = ∠C'
2、解:(1)BE = CF
理由:在△AEB与△AFC中,
∠E = ∠F,∠B = ∠C,AE = AF,
由判定方法“AAS”,得△AEB ≌ △AFC,所以BE = CF.
(2)相等.
理由:因为△AEB ≌ △AFC,
所以∠EAB = ∠FAC,
即∠1 + ∠CAB = ∠2 + ∠CAB,所以∠1 = ∠2.
(3)全等,
理由:因为△AEB ≌ △AFC,
所以AB = AC.
在△ACN与△ABM中,AC = AB,∠C = ∠B,∠BAC是公共角,
由判定方法“ASA”,得△ACN ≌ △ABM
3、3 △ABC ≌ △ADC,△ABE ≌ △ADE,△BCE ≌ △DCE
4、解:相等,
理由:因为CD ⊥ AM,BE ⊥ AM,
所以∠CDM = ∠BEM = 90°.
又因为点M为△ABC边BC的中点,
所以BM = MC.
在△BEM与△CDM中,∠BEM = ∠CDM,∠BME = ∠CMD,BM = CM,
由判定方法“AAS”,得△BEM ≌ △CDM,所以BE = CD.
5、解:如图1 = 4 - 13所示.△ABC就是所求作的三角形.
6、解:能.
由△ABD ≌ △ACE,得∠B = ∠C,AB = AC.AD = AE,
所以AC - AD = AB - AE,
即DC = EB.
又因为∠DCC = ∠EOB,
可由“AAS”判定△OBE ≌ △OCD.
7、解:AC = AE或∠B = ∠D或∠C = ∠E.
因为∠1 = ∠2,所以∠1 + ∠DAC = ∠2 + ∠DAC,
即∠BAC = ∠DAE.
若添加AC = AE,可用“SAS”判定△ABC ≌ △ADE;
若添加∠B = ∠D,可用“ASA”判定△ABC ≌ △ADE;
若添加∠C = ∠E,可用“AAS”判定△ABC ≌ △ADE.
8、解:因为AB∥FC,
所以∠DAE = ∠FCE,∠ADE = ∠CFE
因为E是DF的中点,所以DE = FE,
所以△ADE ≌ △CFE(AAS).
所以AD = CF = 8,BD = AB - AD = 15 - 8 = 7.
9、解:(1)全等.
因为AD = DE,∠ADC = ∠EDB,CD = BD,
可用“SAS”判定△ACD ≌ △EBD.
(2)如图1 - 4 - 14所示.
① 作射线AM,在射线AM上依次截取AD = DE = m.
② 以a,c,2m的长为边作△ABE,使AB = c,BE = b.
③ 连接BD并延长,在BD延长绒上截取线段DC,使DC = BD.
④ 连接AC.△ABC就是所求作的三角形.
10、解:如图1 - 4 - 15所示,满足条件的三角形可作4个.