(1)证明:∵AB=AC,
∴∠B=∠C(对边对等角);
∵AD=AE,
∴∠ADE=∠AED,
∴180°-∠ADE=180°-∠AED,
即∠ADB=∠AEC;
在△ABD和△ACE中,
∠ADB=∠AEC∠B=∠CAB=AC
,
∴△ABD≌△ACE(AAS);
∴BD=CE.
(1)证明:∵AB=AC,
∴∠B=∠C(对边对等角);
∵AD=AE,
∴∠ADE=∠AED,
∴180°-∠ADE=180°-∠AED,
即∠ADB=∠AEC;
在△ABD和△ACE中,
∠ADB=∠AEC∠B=∠CAB=AC
,
∴△ABD≌△ACE(AAS);
∴BD=CE.