题目
8. 已知直三棱柱 ABC--|||-A1B1C1中,侧面AA1B1B为正方形, =BC=2, E,F分-|||-别为AC和CC1的中点, bot (A)_(1)(B)_(1)-|||-(1)求三棱锥 F-EBC 的体积;-|||-(2)已知D为棱A1 B1上的点.证明: bot DE.-|||-A.-|||-D B-|||-C-|||-F-|||-A = B-|||-E-|||-C
题目解答
答案
解析
(1) ∵侧面四边形AA1B1B为正方 形,∴ ${A}_{1}{B}_{1}\bot B{B}_{1}$ , ∵ $BF\bot {A}_{1}{B}_{1}$ 且 $B{B}_{1}\cap BF=B$ ,BB1,BFC平 面BB1C1C, $\therefore {A}_{1}{B}_{1}\bot $ 平面BB1C1C, 又∵ $AB\ykparallel {A}_{1}{B}_{1}$ $\therefore AB\bot $ 平面BB1C1C, 又∵BCC平面BB1 C1C, $\therefore AB\bot BC$, $\because AB=BC=2$ ,E为AC的中点, $\therefore {S}_{\Delta EBC}=\dfrac {1}{2}{S}_{\Delta ABC}=\dfrac {1}{2}\times \dfrac {1}{2}\times 2\times 2=1$ 由直三棱柱知 $CF\bot $ 平面ABC. $\because F$ 为CC1的中点, $\therefore CF=\dfrac {1}{2}C{C}_{1}=\dfrac {1}{2}B{B}_{1}=\dfrac {1}{2}AB=1$ , $\therefore {V}_{F-EBC}=\dfrac {1}{3}{S}_{\Delta EBC}\cdot CF=\dfrac {1}{3}\times 1\times 1=\dfrac {1}{3}$ (2)证明:连接A1E,B1E,
A1 D B1 C F " " A =-= B E C $\because AB=BC$ ,E为AC的中点,∴ $BE\bot AC$. $\because A{A}_{1}\bot $ 平面ABC,BEC平面ABC, $\therefore A{A}_{1}$ $\bot BE$. $\because A{A}_{1}\cap AC=A$ ,AA1,ACC平面AA1C1C, $\therefore BE\bot $ 平面AA1C1C,又A1E C 平 面AA1C1C, $\therefore BE\bot {A}_{1}E$. 在 $Rt\Delta ECF$ 中, $\tan \angle FEC=\dfrac {CF}{CE}=\dfrac {1}{\sqrt {2}}=\dfrac {\sqrt {2}}{2}$
在 $Rt\Delta {A}_{1}AE$ 中, $\tan \angle A{A}_{1}E=\dfrac {AE}{A{A}_{1}}=\dfrac {\sqrt {2}}{2}$, ∴ $\tan \angle FEC=\tan \angle A{A}_{1}E$ , $\therefore \angle FEC$ $=\angle A{A}_{1}E$, $\because \angle A{A}_{1}E+\angle AE{A}_{1}={90}^{\circ }$ , $\therefore \angle FEC+\angle AE{A}_{1}={90}^{\circ }$ , $\therefore \angle {A}_{1}EF={90}^{\circ }$ ,即 ${A}_{1}E\bot EF$. $\because EF\cap EB=E$ ,又EF,EBC平面BEF, $\therefore {A}_{1}E\bot $ 平面BEF,又BFC平面BEF, $\therefore {A}_{1}E\bot BF$ , 又∵ ${A}_{1}{B}_{1}\bot BF$ ${A}_{1}E\cap {A}_{1}{B}_{1}={A}_{1}$ ,A1EC平面 ${A}_{1}{B}_{1}E$ ,A1B1C平面A1B1 E、 $\therefore BF\bot $ 平 面A1B1E. $\because DEC$ 平面A1B1E,∴ $BF\bot DE$.
A1 D B1 C F " " A =-= B E C $\because AB=BC$ ,E为AC的中点,∴ $BE\bot AC$. $\because A{A}_{1}\bot $ 平面ABC,BEC平面ABC, $\therefore A{A}_{1}$ $\bot BE$. $\because A{A}_{1}\cap AC=A$ ,AA1,ACC平面AA1C1C, $\therefore BE\bot $ 平面AA1C1C,又A1E C 平 面AA1C1C, $\therefore BE\bot {A}_{1}E$. 在 $Rt\Delta ECF$ 中, $\tan \angle FEC=\dfrac {CF}{CE}=\dfrac {1}{\sqrt {2}}=\dfrac {\sqrt {2}}{2}$
在 $Rt\Delta {A}_{1}AE$ 中, $\tan \angle A{A}_{1}E=\dfrac {AE}{A{A}_{1}}=\dfrac {\sqrt {2}}{2}$, ∴ $\tan \angle FEC=\tan \angle A{A}_{1}E$ , $\therefore \angle FEC$ $=\angle A{A}_{1}E$, $\because \angle A{A}_{1}E+\angle AE{A}_{1}={90}^{\circ }$ , $\therefore \angle FEC+\angle AE{A}_{1}={90}^{\circ }$ , $\therefore \angle {A}_{1}EF={90}^{\circ }$ ,即 ${A}_{1}E\bot EF$. $\because EF\cap EB=E$ ,又EF,EBC平面BEF, $\therefore {A}_{1}E\bot $ 平面BEF,又BFC平面BEF, $\therefore {A}_{1}E\bot BF$ , 又∵ ${A}_{1}{B}_{1}\bot BF$ ${A}_{1}E\cap {A}_{1}{B}_{1}={A}_{1}$ ,A1EC平面 ${A}_{1}{B}_{1}E$ ,A1B1C平面A1B1 E、 $\therefore BF\bot $ 平 面A1B1E. $\because DEC$ 平面A1B1E,∴ $BF\bot DE$.