【数学A】例題3.3.1：直線と平面の垂直（One More）★★

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% 例題A3.3.1：直線と平面の垂直（One More）★★
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正四面体ABCDについて，次のことを示せ．

(1) 辺ABの中点をMとするとき，辺ABは平面CDMに垂直である．

(2) 辺$\mathrm{BC},\mathrm{AC},\mathrm{AD},\mathrm{BD}$の中点をそれぞれ$\mathrm{P},\mathrm{Q},\mathrm{R},\mathrm{S}$とするとき，四角形PQRSは正方形である．

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% 例題A3.3.1の解答（One More）★★
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(1) $\mathrm{CM},\mathrm{DM}$はそれぞれ，正三角形である$\triangle\mathrm{ABC}$と$\triangle\mathrm{ABD}$の中線であるから，

\[
\mathrm{CM}\perp\mathrm{AB},\mathrm{DM}\perp\mathrm{AB}
\]

よって，辺$\mathrm{AB}$は平面$\mathrm{CDM}$に垂直である．$\blacksquare$

(2) 正四面体の各面の正三角形において，中点連結定理より，

\[
\mathrm{PQ}=\mathrm{QR}=\mathrm{RS}=\mathrm{SP}
\]

また，$\mathrm{AB}\parallel\mathrm{PQ},\mathrm{AB}\parallel\mathrm{RS}$であるから，

\[
\mathrm{PQ}\parallel\mathrm{RS}
\]

したがって，4点$\mathrm{P},\mathrm{Q},\mathrm{R},\mathrm{S}$は同一平面上にある．

さらに，$\mathrm{CD}\parallel\mathrm{QR}$であり，(1)より，$\mathrm{AB}\perp\mathrm{CD}$

ゆえに，$\mathrm{PQ}\perp\mathrm{QR}$，すなわち，$\angle\mathrm{PQR}=90^{\circ}$

よって，各辺の長さが等しく，1つの内角が$90^{\circ}$であるから，四角形$\mathrm{PQRS}$は正方形である．$\blacksquare$

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% 問題A3.3.1：直線と平面の垂直（One More）★★
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四面体$\mathrm{ABCD}$において，$\mathrm{AB}\perp\mathrm{CD},\mathrm{AC}\perp\mathrm{BD}$であるとする．このとき，$\mathrm{A}$から平面$\mathrm{BCD}$に垂線$\mathrm{AH}$を下ろしたとき，点$\mathrm{H}$は$\triangle\mathrm{BCD}$の垂心であることを示せ．

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% 問題A3.3.1の解答（One More）★★
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$\mathrm{AH}\perp$平面$\mathrm{BCD}$より，$\mathrm{AH}\perp\mathrm{CD}$

また，$\mathrm{AB}\perp\mathrm{CD}$より，$\mathrm{CD}$は平面$\mathrm{ABH}$上の交わる2直線$\mathrm{AB},\mathrm{AH}$に垂直であるから，

\[
\mathrm{CD}\perp\text{平面}\mathrm{ABH}
\]

$\mathrm{BH}$は平面$\mathrm{ABH}$上にあるから，$\mathrm{BH}\perp\mathrm{CD}\cdots(\mathrm{i})$

$\mathrm{CD}\perp$平面$\mathrm{ABH}$と同様に考えると，$\mathrm{BD}\perp\text{平面}\mathrm{ACH}$

したがって，$\mathrm{CH}\perp\mathrm{BD}\cdots(\mathrm{ii})$

よって，点$\mathrm{H}$は$\triangle\mathrm{BCD}$の垂心である．$\blacksquare$

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% 例題A3.3.1の側注：直線と平面の垂直（One More）★★
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% 側注1
正三角形の中線は，底辺の垂直二等分線と一致する．

% 側注2
平行な2直線から，平面が定まる．

% 側注3
\begin{align*}
&\mathrm{AB}\parallel\mathrm{PQ},\mathrm{AB}\perp\mathrm{CD}\\
&\Longrightarrow\mathrm{PQ}\perp\mathrm{CD}
\end{align*}

よって，

\begin{align*}
&\mathrm{CD}\parallel\mathrm{QR},\mathrm{PQ}\perp\mathrm{CD}\\
&\Longrightarrow\mathrm{PQ}\perp\mathrm{QR}
\end{align*}

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% 問題A3.3.1の側注：直線と平面の垂直（One More）★★
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% 側注1
直線$l$が，平面$\alpha$上の交わる2直線にそれぞれ垂直なら，$l\perp\alpha$

% 側注2
$\mathrm{AH}\perp\mathrm{BD},\mathrm{AC}\perp\mathrm{BD}$より，$\mathrm{BD}\perp$平面ACH

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% 例題A3.3.1の考え方・補足など：直線と平面の垂直（One More）★★
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% 考え方
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(1) 「直線$h$が平面$\alpha$上の交わる2直線に垂直$\Longrightarrow$直線$h\perp$平面$\alpha$」が成り立つことを利用する．

(2) $\mathrm{PQ}=\mathrm{QR}=\mathrm{RS}=\mathrm{SP}$であることと，1つの内角が$90^{\circ}$であることを示す．そのためには，平行な2直線の一方に垂直な直線は，他方にも垂直であることを利用するとよい．(1)より，$\mathrm{AB}\perp\mathrm{CD}$がわかるから，これと$\mathrm{AB}\parallel\mathrm{PQ},\mathrm{CD}\parallel\mathrm{QR}$より$\mathrm{PQ}\perp\mathrm{QR}$を導くことができる．