Question

如图，点P为△ABC的外心，且=4，=2，则等于    (    )

A.2                B.4                   C.6                 D.8

Answer 1

答案：C  【解析】设∠PAC=α,∠PAB=β,由点P为△ABC的外心,∴|PB|=|PC|,

cosα=(|AP|²+|AC|²-|PC|²),

=(|AP|²+|AB|²-|PB|²),

∴=(|AC|²-|AB|²),故选C.