直三棱柱ABC-A1B1C1的直观图(图1)及三视图(图2)如图所示,D为AC的中点(1)求证:AB1∥平面BDC1(2) 三棱柱ABC-A1B1C1的直观图及三视图(主视图和俯视图是...

\uff082014?\u8d64\u5cf0\u6a21\u62df\uff09\u76f4\u4e09\u68f1\u67f1ABC-A1B1C1\u7684\u76f4\u89c2\u56fe\u53ca\u4e09\u89c6\u56fe\u5982\u56fe\u6240\u793a\uff0cD\u4e3aAC\u7684\u4e2d\u70b9\uff0c\u5219\u4e0b\u5217\u547d\u9898\u662f\u5047\u547d\u9898\u7684\u662f\uff08

\u7531\u4e09\u89c6\u56fe\u77e5\uff1a\u76f4\u4e09\u68f1\u67f1\u7684\u9ad8\u4e3a2\uff0c\u5e95\u9762\u662f\u76f4\u89d2\u8fb9\u957f\u4e3a2\u7684\u7b49\u8170\u76f4\u89d2\u4e09\u89d2\u5f62\uff0c\u2234\u4f53\u79efV=12\u00d72\u00d72\u00d72=4\uff0c\u2234A\u6b63\u786e\uff1b\u7531\u76f4\u4e09\u68f1\u67f1\u7684\u7ed3\u6784\u7279\u5f81\u77e5\uff0c\u68f1\u67f1\u7684\u5e95\u9762\u5468\u957f\u4e3a2+2+22=4+22\uff0c\u2234\u76f4\u4e09\u68f1\u67f1\u7684\u8868\u9762\u79efS=2\u00d712\u00d72\u00d72+\uff084+22\uff09\u00d72=12+42\uff0c\u6545B\u9519\u8bef\uff1b\u53d6A1C1\u4e2d\u70b9O\uff0c\u8fde\u63a5OB1\uff0cAO\uff0c\u2235D\u4e3aAC\u7684\u4e2d\u70b9\uff0c\u2234\u56db\u8fb9\u5f62DAOC1\u4e3a\u5e73\u884c\u56db\u8fb9\u5f62\uff0c\u2234AO\u2225C1D\uff0c\u53c8\u56db\u8fb9\u5f62BDOB1\u4e3a\u5e73\u884c\u56db\u8fb9\u5f62\uff0c\u2234BD\u2225OB1\uff0c\u2234\u5e73\u9762AOB1\u2225\u5e73\u9762BDC1\uff0cAB1?\u5e73\u9762AOB1\uff0c\u2234AB1\u2225\u5e73\u9762BDC1\uff0e\u6545C\u6b63\u786e\uff1b\u2235\u7531\u4e09\u89c6\u56fe\u77e5A1B1\u22a5\u5e73\u9762BCC1B1\uff0cBC1?\u5e73\u9762BCC1B1\uff0c\u2234A1B1\u22a5BC1\uff0cCB1\u22a5BC1\u2234BC1\u22a5\u5e73\u9762A1B1C\uff0c\u2234BC1\u22a5A1C\uff1b\u2235\u7531\u4fa7\u89c6\u56fe\u77e5\u25b3ABC\u4e3a\u7b49\u8170\u76f4\u89d2\u4e09\u89d2\u5f62\uff0cD\u4e3aAC\u7684\u4e2d\u70b9\uff0c\u2234BD\u22a5AC\uff0c\u2234BD\u22a5\u5e73\u9762ACC1A1\uff0c\u2234A1C\u22a5BD\uff0c\u53c8BD\u2229BC1=B\uff0c\u2234A1C\u22a5\u5e73\u9762BDC1\uff0e\u6545D\u6b63\u786e\uff1b\u6545\u9009B\uff0e

\u8fd9\u4e2a\u56fe\u5f62\u5f88\u7279\u6b8a
\u7531
\u4e3b\u89c6\u56fe\u548c\u4fef\u89c6\u56fe\u662f\u6b63\u65b9\u5f62\uff0c\u5de6\u89c6\u56fe\u662f\u7b49\u8170\u76f4\u89d2\u4e09\u89d2\u5f62
\u5f97
BB'C'C\u548cABB'A'\u662f\u6b63\u65b9\u5f62\uff0cABC\u548cA'B'C'\u662f\u7b49\u8170\u76f4\u89d2\u4e09\u89d2\u5f62
\u8981\u6c42\u4e8c\u9762\u89d2A-BC1-D\u7684\u6b63\u5207\u503c
\u7531\u4e8e\u4e8c\u9762\u89d2A-BC1-D\u548c\u4e8c\u9762\u89d2C-BC1-D\u4e92\u4f59
\u4e8e\u662f\u8f6c\u800c\u6c42\u4e8c\u9762\u89d2C-BC1-D\u4f59\u5207
\u5047\u8bbeBC=2
\u8fc7D\u70b9\u4f5cDH\u5782\u76f4BC\u4ea4BC\u4e0e\u70b9H,\u8fde\u63a5BC'\uff0c\u8fc7H\u4f5cHK\u5782\u76f4BC'\u4ea4\u4e0eK\u70b9
\u4e8e\u662fDH=1\uff0cHK=\u4e8c\u5206\u5b50\u6839\u53f7\u4e8c
\u4e8e\u662f
\u4e8c\u9762\u89d2A-BC1-D\u6b63\u5207=\u4e8c\u9762\u89d2C-BC1-D\u4f59\u5207=HK/DH=\u4e8c\u5206\u5b50\u6839\u53f7\u4e8c

(1)证明:取A1C1中点O,连接OB1,AO,
∵D为AC的中点,∴四边形DAOC1为平行四边形,
∴AO∥C1D,又四边形BDOB1为平行四边形,
∴BD∥OB1,∴平面AOB1∥平面BDC1,AB1?平面AOB1
∴AB1∥平面BDC1
(2)证明:∵由三视图知A1B1⊥平面BCC1B1,BC1?平面BCC1B1
∴A1B1⊥BC1,CB1⊥BC1
∴BC1⊥平面A1B1C,∴BC1⊥A1C;
∵由侧视图知△ABC为等腰直角三角形,D为AC的中点,
∴BD⊥AC,∴BD⊥平面ACC1A1
∴BD⊥AC1
(3)解:由三视图知:直三棱柱的高为2,
底面是直角边长为2的等边三角形,
∴体积V=
1
2
×2×2×2=4.

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