【数学A】例題3.1.2：三角形の性質（One More）★★

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% 例題A3.1.2：三角形の性質（One More）★★
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$\triangle\mathrm{ABC}$において，辺$\mathrm{BC}$の中点を$\mathrm{M}$とし，$\angle\mathrm{AMB},\angle\mathrm{AMC}$の二等分線が辺$\mathrm{AB},\mathrm{AC}$と交わる点をそれぞれ$\mathrm{D}$，$\mathrm{E}$とする．このとき，$\mathrm{DE}\parallel\mathrm{BC}$であることを示せ．

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% 例題A3.1.2の解答（One More）★★
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$\triangle\mathrm{AMB}$に注目すると，MDは$\angle\mathrm{AMB}$の二等分線であるから，

\[
\mathrm{MA}:\mathrm{MB}=\mathrm{AD}:\mathrm{DB}\cdots(\mathrm{i})
\]

同様に，$\triangle\mathrm{AMC}$に注目すると，MEは$\angle\mathrm{AMC}$の二等分線であるから，

\[
\mathrm{MA}:\mathrm{MC}=\mathrm{AE}:\mathrm{EC}\cdots(\mathrm{ii})
\]

$\mathrm{AM}$は$\triangle\mathrm{ABC}$の中線であるから，$\mathrm{MB}=\mathrm{MC}$

したがって，(i)，(ii)より，

\[
\mathrm{AD}:\mathrm{DB}=\mathrm{MA}:\mathrm{MB}=\mathrm{AE}:\mathrm{EC}
\]

よって，$\mathrm{AD}:\mathrm{DB}=\mathrm{AE}:\mathrm{EC}$より，$\mathrm{DE}\parallel\mathrm{BC}\blacksquare$

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% 問題A3.1.2：三角形の性質（One More）★★
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$\triangle\mathrm{ABC}$において，辺$\mathrm{BC}$の中点を$\mathrm{M}$とし，$\angle\mathrm{AMB},\angle\mathrm{AMC}$の二等分線が辺$\mathrm{AB},\mathrm{AC}$と交わる点をそれぞれ$\mathrm{D}$，$\mathrm{E}$とする．このとき，$\mathrm{DE}<\mathrm{BD}+\mathrm{CE}$であることを示せ．

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% 問題A3.1.2の解答（One More）★★
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右の図のように，半直線$\mathrm{MA}$上で，$\mathrm{BM}=\mathrm{CM}=\mathrm{FM}$となるように点$\mathrm{F}$をとる．

$\triangle\mathrm{BDM}$と$\triangle\mathrm{FDM}$において，2組の辺とその間の角がそれぞれ等しいから，$\triangle\mathrm{BDM}\equiv\triangle\mathrm{FDM}$

したがって，$\mathrm{BD}=\mathrm{FD}\cdots(\mathrm{i}),\angle\mathrm{DBM}=\angle\mathrm{DFM}\cdots(\mathrm{ii})$

$\triangle\mathrm{CEM}$と$\triangle\mathrm{FEM}$においても同様に考えると，$\triangle\mathrm{CEM}\equiv\triangle\mathrm{FEM}$

ゆえに，$\mathrm{CE}=\mathrm{FE}\cdots(\mathrm{iii}),\angle\mathrm{ECM}=\angle\mathrm{EFM}\cdots(\mathrm{iv})$

(ii)，(iv)より，

\begin{align*}
\angle\mathrm{DFM}+\angle\mathrm{EFM}&=\angle\mathrm{DBM}+\angle\mathrm{ECM}\\
&=\angle\mathrm{ABC}+\angle\mathrm{ACB}\\
&=180^{\circ}-\angle\mathrm{BAC}<180^{\circ}
\end{align*}

したがって，3点$\mathrm{D},\mathrm{F},\mathrm{E}$は同一直線上にない．

ゆえに，三角形の成立条件より，$\mathrm{DE}<\mathrm{FD}+\mathrm{FE}\cdots(\mathrm{v})$

よって，(i)，(iii)，(v)より，$\mathrm{DE}<\mathrm{BD}+\mathrm{CE}\blacksquare$

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% 問題A3.1.2の側注：三角形の性質（One More）★★
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% 側注1
2つの線分の長さの和は，1つの線分の長さより大きいことを示すことから，三角形の成立条件「三角形の2辺の長さの和は，残りの辺の長さより大きい」を用いることを考える．$\mathrm{BD}=\mathrm{FD},\mathrm{CE}=\mathrm{FE}$となる$\triangle\mathrm{FDE}$が存在することを示すために，半直線$\mathrm{MA}$上で，$\mathrm{BM}=\mathrm{CM}=\mathrm{FM}$となるように点$\mathrm{F}$をとる．

% 側注2
3点が同一直線上にあるとき，$\mathrm{DE}=\mathrm{BD}+\mathrm{CE}$となり，$\triangle\mathrm{FDE}$が存在せず，三角形の成立条件を適用できない．

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% 例題A3.1.2の考え方・補足など：三角形の性質（One More）★★
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% 考え方
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$\mathrm{DE}\parallel\mathrm{BC}$を示すことから，$\triangle\mathrm{ABC}$において，次のような平行線と線分の比の性質を利用する．

\[
\mathrm{DE}\parallel\mathrm{BC}\Longleftrightarrow\mathrm{AD}:\mathrm{DB}=\mathrm{AE}:\mathrm{EC}
\]

そこで，$\mathrm{AD}:\mathrm{DB}=\mathrm{AE}:\mathrm{EC}$を導くために，$\triangle\mathrm{AMB},\triangle\mathrm{AMC}$のそれぞれに注目して考える．

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平行線と線分の比の性質を用いて，平行であることを導く．