問題の解答
検索用コード（LaTeX）
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% 例題A3.2.5：方べきの定理の逆（One More）★★
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鋭角三角形$\mathrm{ABC}$の頂点$\mathrm{A}$から辺$\mathrm{BC}$に垂線$\mathrm{AP}$を引き，$\mathrm{P}$から辺$\mathrm{AB}$，$\mathrm{AC}$に垂線を下ろし，それぞれの交点を$\mathrm{Q}$，$\mathrm{R}$とする．このとき，4点$\mathrm{B},\mathrm{C},\mathrm{R},\mathrm{Q}$は同一円周上にあることを示せ．
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% 例題A3.2.5の解答（One More）★★
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$\angle\mathrm{BQP}=90^{\circ}$より，$\triangle\mathrm{BPQ}$は$\mathrm{BP}$を直径とする円に内接する．
また，$\angle\mathrm{APB}=90^{\circ}$であるから，$\mathrm{AP}$はこの円の接線である．
したがって，方べきの定理より，
\[
\mathrm{AP}^2=\mathrm{AQ}\cdot\mathrm{AB}\cdots(\mathrm{i})
\]
同様に，$\angle\mathrm{CRP}=90^{\circ}$，$\angle\mathrm{APC}=90^{\circ}$であるから，
\[
\mathrm{AP}^2=\mathrm{AR}\cdot\mathrm{AC}\cdots(\mathrm{ii})
\]
(i)，(ii)より，$\mathrm{AQ}\cdot\mathrm{AB}=\mathrm{AR}\cdot\mathrm{AC}$
よって，方べきの定理の逆より，4点$\mathrm{B},\mathrm{C},\mathrm{R},\mathrm{Q}$は同一円周上にある．$\blacksquare$
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% 例題A3.2.5の別解
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$\triangle\mathrm{APB}$と$\triangle\mathrm{AQP}$について，$\angle\mathrm{BAP}=\angle\mathrm{QAP},\angle\mathrm{APB}=\angle\mathrm{AQP}=90^{\circ}$
したがって，$\triangle\mathrm{APB}\backsim\triangle\mathrm{AQP}$
ゆえに，$\mathrm{AP}:\mathrm{AQ}=\mathrm{AB}:\mathrm{AP}$より，
\[
\mathrm{AP}^2=\mathrm{AQ}\cdot\mathrm{AB}\cdots(\mathrm{i})
\]
同様に，$\triangle\mathrm{APC}$と$\triangle\mathrm{ARP}$において，
\[
\angle\mathrm{CAP}=\angle\mathrm{RAP},\angle\mathrm{APC}=\angle\mathrm{ARP}=90^{\circ}
\]
したがって，$\triangle\mathrm{APC}\backsim\triangle\mathrm{ARP}$
ゆえに，$\mathrm{AP}:\mathrm{AR}=\mathrm{AC}:\mathrm{AP}$より，$\mathrm{AP}^2=\mathrm{AR}\cdot\mathrm{AC}\cdots(\mathrm{ii})$
(i)，(ii)より，$\mathrm{AQ}\cdot\mathrm{AB}=\mathrm{AR}\cdot\mathrm{AC}$
よって，方べきの定理の逆より，4点$\mathrm{B},\mathrm{C},\mathrm{R},\mathrm{Q}$は同一円周上にある．$\blacksquare$
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% 問題A3.2.5：方べきの定理の逆（One More）★★
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右の図のように，2つの円$\mathrm{O},\mathrm{O}^{\prime}$が2点$\mathrm{Q},\mathrm{R}$で交わっており，QRの延長上の点Pから，円$\mathrm{O},\mathrm{O}^{\prime}$にそれぞれ$\mathrm{A},\mathrm{B}$および$\mathrm{C},\mathrm{D}$で交わる直線を引くとする．このとき，4点A，B，C，Dは同一円周上にあることを示せ．
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% 問題A3.2.5の解答（One More）★★
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円Oにおいて，方べきの定理より，$\mathrm{PA}\cdot\mathrm{PB}=\mathrm{PQ}\cdot\mathrm{PR}$
円$\mathrm{O}^\prime$において，方べきの定理より，$\mathrm{PC}\cdot\mathrm{PD}=\mathrm{PQ}\cdot\mathrm{PR}$
したがって，$\mathrm{PA}\cdot\mathrm{PB}=\mathrm{PC}\cdot\mathrm{PD}$
よって，方べきの定理の逆より，4点A，B，C，Dは同一円周上にある．$\blacksquare$
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