在三棱柱ABC-A1B1C1中，底面ABC是等边三角形，D为AB的中点，求BC1平行平面A1CD
\u5982\u56fe\uff0c\u5728\u4e09\u68f1\u67f1ABC\u00b7A1B1C1\u4e2d\uff0cE,F\u5206\u522b\u662fAB,AC\u7684\u4e2d\u70b9\uff0c\u5e73\u9762EB1C1F\u5c06\u4e09\u68f1\u67f1\u5206\u6210\u4e24\u90e8\u5206\uff0c\u5ef6\u957fA1A,B1E,C1F\u4ea4\u4e8e\u70b9P,
\u8bbe\u539f\u4e09\u68f1\u67f1\u5e95\u9762\u79ef\u4e3aS,\u9ad8\u4e3ah .
\u5219\u4e09\u68f1\u9525P-A1B1C1\u7684\u4f53\u79ef\u4e3a\uff1a1/3*S*2h=2/3*Sh
\u5c0f\u4e09\u68f1\u9525P-ABC\u7684\u4f53\u79ef\u4e3a\uff1a1/3*1/4S*h=1/12*Sh
\u6240\u4ee5\uff1aV1=2/3*Sh-1/12*Sh=7/12Sh
V2=Sh-7/12Sh=5/12Sh
\u6240\u4ee5\uff1a
V1:V2=7:5
\u6216
\u8bbe\u4ee5AA1\u4e3a\u6395\u7684\u90e8\u5206\u4f53\u79ef\u4e3aV1.\u22bfABC\u9762\u79ef\uff1dS,\u5219\u22bfAEF\u9762\u79ef\uff1dS\uff0f4.
\u518d\u8bbe\u4e09\u68f1\u67f1\u7684\u9ad8\u4e3aH.\u6309\u6395\u53f0\u516c\u5f0f\u5f97\uff1a
V1\uff1d\uff081/3\uff09H\uff08S+\uff08S/4\uff09+\u221a[S\uff08S/4\uff09]\uff1d\uff087/12\uff09HS.
V2\uff1dHS-\uff087/12\uff09HS\uff1d\uff085/12\uff09HS.
\u2234V1\u2236V2\uff1d7\u22365.
\u8fde\u63a5AC1\uff0c\u4e0e\u76f4\u7ebfA1C\u4ea4\u70b9\u4e3aO\uff0c\u8fde\u63a5OD\u3002
\u53c8\u56e0\u4e3a\u662f\u4e09\u68f1\u67f1ABC-A1B1C1
\u6240\u4ee5AO=OC1
\u56e0\u4e3aD\u4e3aAB\u7684\u4e2d\u70b9
\u6240\u4ee5\u5728\u4e09\u89d2\u5f62AC1B\u4e2d\uff0cOD\u4e3a\u4e2d\u4f4d\u7ebf\uff0c
\u6240\u4ee5OD\u5e73\u884c\u4e8eBC1
∵D为AB的中点，
∴DO∥BC1
∵BC1�6�5平面A1CD，DO�6�3平面A1CD，
∴BC1∥平面A1CD；
（Ⅱ）∵等边△ABC，D为AB的中点，
∴CD⊥AB
∵CD⊥DA1，DA1∩AB=D，
∴CD⊥平面ABB1A1
∵BB1�6�3平面ABB1A1
∴BB1⊥CD，
∵矩形BCC1B1
∴BB1⊥BC
∵BC∩CD=C∴BB1⊥平面ABC
∵底面△ABC是等边三角形，
∴三棱柱ABC-A1B1C1是正三棱柱．
绛旓細鈭粹柍A1B1C1鏄瓑鑵癛T鈻筹紝鈭碆1C1=鈭2锛屸埓DC1=鈭3锛孋1D^2+AD^2=5=AC1^2锛屾牴鎹嬀鑲″畾鐞嗛嗗畾鐞嗭紝鈻矯1DA鏄疪T鈻筹紝鈭碈1D鈯D锛屸埖C1D鈭〢D=D锛屸埓A1D鈯ュ钩闈1DC1銆傛柦涓伙紝鎴戠湅浣犻楠兼竻濂囷紝鍣ㄥ畤杞╂槀,涓旀湁鎱ф牴,涔冩槸涓囦腑鏃犱竴鐨勬鏋楀鎵.娼滃績淇範,灏嗘潵蹇呮垚澶у櫒锛岄剻浜烘湁涓皬灏忕殑鑰冮獙璇风偣鍑诲湪涓嬬瓟妗堟梺鐨...
绛旓細璇佹槑锛氳繛鎺B1C锛屼笌BC1浜や簬O鐐癸紝鍒欐湁O鏄疊1C鐨勪腑鐐癸紟鍙圖鏄疉C鐨勪腑鐐癸紝鍒欐湁OD锛忥紡AB1锛嶰D鍦ㄩ潰BDC1鍐咃紝鍒欐湁AB1锛忥紡闈DC1锛2锛庤繃B浣淏E鍨傜洿浜嶢C 鐢变簬AA1鍨傜洿浜庨潰ABC锛屽垯鏈堿A1鍨傜洿浜嶣E锛屾墍浠ユ湁BE鍨傜洿浜庨潰AA1C1C 鍗矪E鏄簳闈A1C1D鐨勯珮锛庡張锛欰A1锛滱B锛2锛孊C锛3锛庢墍浠C锛濇牴鍙凤紙2锛2锛3锛2...
绛旓細鎵浠C鈭B1涓擡C=12BB1锛屾墍浠C鈭D涓擡C=OD锛庢墍浠ワ紝鍥涜竟褰COD涓哄钩琛屽洓杈瑰舰锛庢墍浠O鈭D锛庡張CD?骞抽潰A1BE锛孍O?骞抽潰A1BE锛屽垯CD鈭ュ钩闈1BE锛涳紙2锛夊洜涓鍦ㄤ笁妫辨煴ABC-A1B1C1涓锛屾瘡涓晶闈㈠潎涓烘鏂瑰舰锛孌涓哄簳杈笰B鐨勪腑鐐癸紝鎵浠D鈯B锛孋D鈯1A锛屽洜涓篈1A鈭〢B=A锛屾墍浠D鈯ュ钩闈1ABB1锛
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绛旓細瑙ｏ細鐩涓夋１鏌盇BC-A1B1C1涓锛孉B=BC=2锛孊B1=2锛屸垹ABC=90掳锛孍锛孎鍒嗗埆涓篈A1锛孊1C1鐨勪腑鐐癸紝鎵浠1E=1锛孉1E鈯F锛孍F=2锛屽洓闈綋涓篊-A1EF鐨勯珮涓猴細BC=2锛庘埓鍥涢潰浣撲负C-A1EF鐨勪綋绉负锛13脳12脳1脳2脳2=13锛庢晠绛旀涓猴細13锛
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