题目内容
已知P为△ABC所在平面外一点,G1、G2、G3分别是△PAB、△PCB、△PAC的重心.
(1)求证:平面G1G2G3∥平面ABC;
(2)求S△
∶S△ABC.
(1)求证:平面G1G2G3∥平面ABC;
(2)求S△
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(1)证明略(2)S△
∶S△ABC=1∶9
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(1) 如图所示,连接PG1、PG2、PG3并延长分别与边AB、BC、AC交于点D、E、F,连接DE、EF、FD,则有PG1∶PD=2∶3,
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PG2∶PE=2∶3,∴G1G2∥DE.
又G1G2不在平面ABC内,
∴G1G2∥平面ABC.同理G2G3∥平面ABC.
又因为G1G2∩G2G3=G2,
∴平面G1G2G3∥平面ABC.
(2) 由(1)知
=
,∴G1G2=
DE.
又DE=
AC,∴G1G2=
AC.
同理G2G3=
AB,G1G3=
BC.
∴△G1G2G3∽△CAB,其相似比为1∶3,
∴S△
∶S△ABC=1∶9.
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PG2∶PE=2∶3,∴G1G2∥DE.
又G1G2不在平面ABC内,
∴G1G2∥平面ABC.同理G2G3∥平面ABC.
又因为G1G2∩G2G3=G2,
∴平面G1G2G3∥平面ABC.
(2) 由(1)知
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又DE=
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同理G2G3=
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∴△G1G2G3∽△CAB,其相似比为1∶3,
∴S△
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