如图s是△ABC所在平面外面外一点,SA⊥SB,sB⊥SC,SC⊥SA,H是△ABc的垂心,求证：S

如图s是△ABC所在平面外面外一点,SA⊥SB,sB⊥SC,SC⊥SA,H是△ABc的垂心,求证：S

∵SC⊥SA,SC⊥SB,SA∩SB=S,∴SC⊥平面SAB,∵AB∈平面SAB,∴SC⊥AB,连结CH,延长交AB于D,∵H是△ABCC垂心,∴CD⊥AB,SC∩CD=C,∴AB⊥平面SDC,SH∈平面SCD,∴SH⊥AB,同理BC⊥SH,AB∩BC=B,∴SH⊥平面ABC.

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