(1)∵CM⊥AM,∠DCM=α,
∴∠CDM=∠ADB=∠B=90°-α,
∴∠BAD=180°-2∠ABD=180°-2(90°-α)=2α;
(2)延长AM到F使MF=AM,则有AC=CF
∵AD平分∠CAB
∴∠CAF=∠BAF=∠F
∴CF∥AB
∴∠FCD=∠ABD=∠ADB=∠CDF
∴CF=DF
∵AD+DF=2MA
∴AB+AC=2MA
(1)∵CM⊥AM,∠DCM=α,
∴∠CDM=∠ADB=∠B=90°-α,
∴∠BAD=180°-2∠ABD=180°-2(90°-α)=2α;
(2)延长AM到F使MF=AM,则有AC=CF
∵AD平分∠CAB
∴∠CAF=∠BAF=∠F
∴CF∥AB
∴∠FCD=∠ABD=∠ADB=∠CDF
∴CF=DF
∵AD+DF=2MA
∴AB+AC=2MA