D
分析:根據(jù)SSS證△BAD≌△CAE,推出∠ABD=∠ACE,∠ADB=∠AEC,∠BAD=∠CAE,求出∠BAC=∠EAD,根據(jù)等腰三角形性質(zhì)求出∠ABC=∠ACB,∠ADE=∠AED,根據(jù)三角形內(nèi)角和定理得出∠ABC=∠ACB=∠ADE=∠AED(共4對),即可得出答案.
解答:
解:∵在△BAD和△CAE中,
,
∴△BAD≌△CAE(SSS),
∴∠ABD=∠ACE,∠ADB=∠AEC,∠BAD=∠CAE,
∴∠BAD+∠DAC=∠CAE+∠DAC,
∴∠BAC=∠EAD,
∵AB=AC,AD=AE,
∴∠ABC=∠ACB,∠ADE=∠AED,
∵∠BAC=∠DAE,∠BAC+∠ABC+∠ACB=180°,∠DAE+∠ADE+∠AED=180°,
∴∠ABC=∠ACB=∠ADE=∠AED,
即相等的角有∠BAD=∠CAE,∠ABD=∠ACE,∠ADB=∠AEC,∠BAC=∠DAE,∠ABC=∠ACB,∠ADE=∠AED,∠ADE=∠ABC,∠ADE=∠ACB,∠AED=∠ABC,∠AED=∠ACB,共10對.
故選D.
點評:本題考查了等腰三角形性質(zhì),全等三角形性質(zhì)和判定,三角形的內(nèi)角和定理等知識點的綜合運用,能做到不重不漏是解此題的關(guān)鍵.