C于D,ADP90,
DPO90.
PD是⊙O的切线.
(2)连结AP,
AB是直径,
CPDAB
APB90,
ABAC2,CAB120,
O
BAP60.
BP3,BC23.
28.(10分)解:(1)BMDNMN成立.)如图,把△AND绕点A顺时针90,得到△ABE,则可证得E,B,M三点共线(图形画正确)(3分)证明过程中,证得:EAMNAM证得:△AEM≌△ANM)AD
NBMC
MEMNMEBEBMDNBM
E
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DNBMMN(2)DNBMMN
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