证明:
过A作AE垂直BC,过D作DF垂直BC于F,则BE=CF,如图
于是EF=BC
AC²=AE²+(BC-CF)²
BD²=DF²+(BC+CF)²
∵ AE=DF,BE=CF
∴ AC²+BD²=2AE²+2BC²+2BE²=2(AE²+BE²)+2BC²=2AB²+2BC²
即AC²+BD²=2(AB²+BC)²
证明:
过A作AE垂直BC,过D作DF垂直BC于F,则BE=CF,如图
于是EF=BC
AC²=AE²+(BC-CF)²
BD²=DF²+(BC+CF)²
∵ AE=DF,BE=CF
∴ AC²+BD²=2AE²+2BC²+2BE²=2(AE²+BE²)+2BC²=2AB²+2BC²
即AC²+BD²=2(AB²+BC)²