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$解:(1)∠BCE與∠BAC相等或互補,理由如下:$

$當(dāng)D在線段BC上時,如圖1(a)$

$由題,AD=AE$

$∵∠DAE=∠BAC,∴∠BAD=∠CAE$

$在△ABD和△ACE中$

${{\begin{cases} {{AB=AC}} \\ {∠BAD=∠CAE} \\ {AD=AE} \end{cases}}}$

$∴△ABD≌△ACE(SAS)$

$∴∠ABC=∠ACE$

$∴∠BCE=∠BCA+∠ACE=∠BCA+∠ABC$

$而∠BAC+∠BCA+∠ABC=180°$

$∴∠BCE+∠ABC=180°$

$當(dāng)D在BC延長線上時,如圖1(b)$

$同理可證△ABD≌△ACE,∴∠ABD=∠ACE$

$∴∠BCE+∠ABC=180°$

$當(dāng)D在CB延長線上時,如圖1(c)$

$同理可證△ABD≌△ACE,∴∠ABD=∠ACE$

$∵∠ABD=∠BAC+∠ACB,∠ACE=∠ACB+∠BCE$

$∴∠BCE=∠BAC$

$(2)（更多請點擊查看作業(yè)精靈詳解）$

$解:∠EBC與∠BAC相等或互補,理由:$

$當(dāng)D在線段BC上時,如圖2(a)$

$∵∠DAE=∠BAC,∴∠CAD=∠BAE$

$在△ACD和△ABE中$

${{\begin{cases} {{AC=AB}} \\ {∠CAD=∠BAE} \\ {AD=AE} \end{cases}}}$

$∴△ACD≌△ABE(SAS)$

$∴∠ACD=∠ABE$

$∵∠EBC=∠ABE+∠ABC$

$∴∠EBC=∠ABC+∠ACB$

$而∠BAC+∠BCA+∠ABC=180°$

$∴∠EBC+∠BAC=180°$

$當(dāng)D在BC延長線上時,如圖2(b)$

$同理可證△ABE≌△ACD,∴∠ABE=∠ACD$

$∵∠ABE=∠ABC+∠EBC,∠ACD=∠ABC+∠BAC$

$∴∠EBC=∠BAC$

$當(dāng)D在CB延長線上時,如圖2(c)$

$同理可證△ABE≌△ACD,∴∠ABE=∠ACD$

$∵∠EBC=∠ABE+∠ABC$

$∴∠EBC=∠ABC+∠ACB$

$而∠BAC+∠BCA+∠ABC=180°$

$∴∠EBC+∠BAC=180°$

$綜上,∠EBC與∠BAC相等或互補$

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