
解:在BC上截取一点F,使BF=BA,连结FD ∵∠ABC=40° AB=AC∴∠DCF=40°∠BAD=100° ∠ADB=60° ∵BD平分∠ABC ∴∠ABD=∠FBD=20° ∴△ABD与△FBD全等(ASA)∴AD=FD ∠FDB=∠ADB=60° ∵DE=AD ∠CDE=∠ADB=∠FDB=60°=∠CDF ∴FD=ED ∴△CDE与△CDB全等 (SAS)∴∠ECA=∠DCF=40°

解:在BC上截取一点F,使BF=BA,连结FD ∵∠ABC=40° AB=AC∴∠DCF=40°∠BAD=100° ∠ADB=60° ∵BD平分∠ABC ∴∠ABD=∠FBD=20° ∴△ABD与△FBD全等(ASA)∴AD=FD ∠FDB=∠ADB=60° ∵DE=AD ∠CDE=∠ADB=∠FDB=60°=∠CDF ∴FD=ED ∴△CDE与△CDB全等 (SAS)∴∠ECA=∠DCF=40°