18.(12分)-|||-如图,四面体ABC D中, bot CD , =CD,-|||-angle ADB=angle BDC, E为AC的中点.-|||-(1)证明:平面 bot 平面ACD;-|||-(2)设 AB=BD=2 . angle ACB=(60)^circ , 点F在BD上;-|||-当 Delta AFC 的面积最小时,求CF与平面ABD所成的角的正-|||-弦值.-|||-D-|||-F-|||-1-|||-B-|||-A
题目解答
答案
【答案】
$\left(1\right)$见解析；$\left(2\right)$$\dfrac{4\sqrt{3}}{7}$
【解析】
$\left(1\right)$$\because AD=CD$，$E$为$AC$的中点，
$\therefore DE\bot AC$.
$\because AD=CD$，$\angle ADB=\angle BDC$，$BD=BD$，
$\therefore \triangle ABD\ykcong \triangle CBD$，
$\therefore AB=CB$.
又$\because $$E$为$AC$的中点，
$\therefore BE\bot AC$.
又$\because DE\cap BE=E$，$DE$，$BE\subset $平面$BED$，
$\therefore AC\bot $平面$BED$.
$\because AC\subset $平面$ACD$，
$\therefore $平面$ACD\bot $平面$BED$.
$\left(2\right)$$\because AD=CD$，$\angle ADB=\angle BDC$，$DF=DF$，
$\therefore \triangle AFD\ykcong \triangle CFD$，
$\therefore AF=CF$.
连接$EF$.
$\because E$为$AC$的中点，
$\therefore EF\bot AC$.
由$\left(1\right)$可知$AB=CB$.
又$\because \angle ACB={60}^{\circ }$，$AB=2$，
$\therefore \triangle ABC$为边长为$2$的等边三角形，
$\therefore AC=2$，
$BE=AB\sin {60}^{\circ }=2\times \dfrac{\sqrt{3}}{2}=\sqrt{3}$.
$\because AD\bot CD$，$E$为$AC$的中点，
$\therefore DE=\dfrac{1}{2}AC=\dfrac{1}{2}\times 2=1$，
$\therefore D{E}^{2}+B{E}^{2}=B{D}^{2}$，
$\therefore DE\bot BE$，
$\therefore {S}_{\triangle AFC}=\dfrac{1}{2}AC\cdot EF=\dfrac{1}{2}\times 2\cdot EF=EF$.
要使得$\triangle AFC$的面积最小，只需$EF$最小，即$EF\bot BD$.
此时在$Rt\triangle BED$中，$EF=\dfrac{BE\cdot DE}{BD}=\dfrac{\sqrt{3}\times 1}{2}=\dfrac{\sqrt{3}}{2}$，
则$DF=\sqrt{D{E}^{2}-E{F}^{2}}=\sqrt{{1}^{2}-{\left(\dfrac{\sqrt{3}}{2}\right)}^{2}}=\dfrac{1}{2}$，
故$DF=\dfrac{1}{4}DB$.
$\because DE\bot AC$，$DE\bot BE$，$AC\cap BE=E$，$AC$，$BE\subset $平面$ABC$，
$\therefore DE\bot $平面$ABC$.
以$E$为坐标原点，$EA$，$EB$，$ED$所在直线分别为$x$轴，$y$轴，$z$轴，
建立如图所示空间直角坐标系，
则$A\left(1,0,0\right)$，$B\left(0,\sqrt{3},0\right)$，$C\left(-1,0,0\right)$，$D\left(0,0,1\right)$，
则$\overrightarrow{AB}=\left(-1,\sqrt{3},0\right)$，$\overrightarrow{AD}=\left(-1,0,1\right)$.
设平面$ABD$的法向量为$\overrightarrow{a}=\left(x,y,z\right)$，
则$\left\{\begin{array}{l}\overrightarrow{AB}\cdot \overrightarrow{a}=-x+\sqrt{3}y=0\\ \overrightarrow{AD}\cdot \overrightarrow{a}=-x+z=0\end{array}\right.$
不妨取$y=1$，则$x=z=\sqrt{3}$，故$\overrightarrow{a}=\left(\sqrt{3},1,\sqrt{3}\right)$.
$\overrightarrow{CF}=\overrightarrow{CD}+\overrightarrow{DF}=\overrightarrow{CD}+\dfrac{1}{4}\overrightarrow{DB}$
$=\left(1,0,1\right)+\dfrac{1}{4}\left(0,\sqrt{3},-1\right)=\left(1,\dfrac{\sqrt{3}}{4},\dfrac{3}{4}\right)$.
设$CF$与平面$ABD$所成角的大小为$\theta $，
则$\sin \theta =\left|\cos \lt \overrightarrow{CF},\overrightarrow{a}\gt \right|=\dfrac{\left|\overrightarrow{CF}\cdot \overrightarrow{a}\right|}{\left|\overrightarrow{CF}\right|\cdot \left|\overrightarrow{a}\right|}$
$=\dfrac{\left|1\times \sqrt{3}+\frac{\sqrt{3}}{4}\times 1+\frac{3}{4}\times \sqrt{3}\right|}{\sqrt{{1}^{2}+{\left(\frac{\sqrt{3}}{4}\right)}^{2}+{\left(\frac{3}{4}\right)}^{2}}\times \sqrt{{\left(\sqrt{3}\right)}^{2}+{1}^{2}+{\left(\sqrt{3}\right)}^{2}}}$
$=\dfrac{4\sqrt{3}}{7}$，
即$CF$与平面$ABD$所成角的正弦值为$\dfrac{4\sqrt{3}}{7}$.