解:1、延长BA到F,使AF=AC,连接EF则
AB+AC=AB+AF=BF=BE
∴∠BEF=∠F
设∠B=X∵已知∠BAD=∠DAC=9∴∠ACE=x+18 =∠F
∵AD⊥AE∴∠CAE=81 = ∠FAE
易证⊿CAE≌⊿FAE∴∠ACE=∠F=X+18
在⊿ABE中∠AEB=180-99-X=81-X=∠FEA
∵∠BEF=∠F ∴2(81-X)=X+18 X=48 即∠B=48
2、不会是这个吧AM=MN=NB=BN
(1)延长BA到F时AF=AC,连接EF,则
BF=AB+AF=AB+AC=BE
∠F=∠BEF
∠CAE=90°-9°=81°
∠FAE=180°-81°-18°=81°
且AF=AC,AE为公共边。可知⊿CAE≌⊿FAE
∠F=∠ACE
在四边形ACFE中
∠FAC=2*81°=162°
∠F=∠ACE=∠FEC=1/3(360°-162°)=66°
∠B=∠ACE-∠BAC=66°-18°=48°
(1)∠B=48°