解证明题
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解证明题
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证明:(1)∵BD⊥直线m,CE⊥直线m,
∴∠BDA=∠CEA=90°,
∵∠BAC=90°,
∴∠BAD+∠CAE=90°,
∵∠BAD+∠ABD=90°,
∴∠CAE=∠ABD,
∵在△ADB和△CEA中
{∠ABD=∠CAE
{∠BDA=∠CEA
{AB=AC
∴△ADB≌△CEA(AAS),
∴AE=BD,AD=CE,
∴DE=AE+AD=BD+CE;
(2)∵∠BDA=∠BAC=α,
∴∠DBA+∠BAD=∠BAD+∠CAE=180°-α,
∴∠CAE=∠ABD,
∵在△ADB和△CEA中
{∠ABD=∠CAE
{∠BDA=∠CEA
{AB=AC
∴△ADB≌△CEA(AAS),
∴AE=BD,AD=CE,
∴DE=AE+AD=BD+CE
∴∠BDA=∠CEA=90°,
∵∠BAC=90°,
∴∠BAD+∠CAE=90°,
∵∠BAD+∠ABD=90°,
∴∠CAE=∠ABD,
∵在△ADB和△CEA中
{∠ABD=∠CAE
{∠BDA=∠CEA
{AB=AC
∴△ADB≌△CEA(AAS),
∴AE=BD,AD=CE,
∴DE=AE+AD=BD+CE;
(2)∵∠BDA=∠BAC=α,
∴∠DBA+∠BAD=∠BAD+∠CAE=180°-α,
∴∠CAE=∠ABD,
∵在△ADB和△CEA中
{∠ABD=∠CAE
{∠BDA=∠CEA
{AB=AC
∴△ADB≌△CEA(AAS),
∴AE=BD,AD=CE,
∴DE=AE+AD=BD+CE