Re: [中學] 奧數
6)
△AMP ~ △DMA (AA~) => ∠ADM = ∠PAM
延長AP交BC於E => △AEB = △DMA (ASA)
=> BE = AM = AN
=> ∠END = 90 = ∠NEC
∠DNC = ∠DEC
PNDCE共圓 => ∠NPD = ∠NED
=> ∠APN = 90 - ∠NPD = 90 - ∠NED
= ∠DEC = ∠DNC
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