全部!​

全部!​
全部!​

全部!​
正如提示,过E作EF⊥AB于F,
(1)∵∠FAE=∠DAE,∠ADE=∠AFE=90°,=>∠AEF=∠AED,
=>△AEF≌△AED(ASA),=>ED=EF,
同理,=>△BEF≌△BEC(ASA),=>EC=EF,
=>ED=EC,
(2)由(1)知∠AEF=∠AED,∠BEF=∠BED,
又∠AEF+∠AED+∠BEF+∠BED=180°,
=>∠AEB=∠AEF+∠BEF= 90°,
(3)由△AEF≌△AED(ASA)=>AD=AF,
由△BEF≌△BEC(ASA)=>BC=BF,
=>AD+BC=AF+BF=AB.
祝愉快