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数学 中学生

問題8の(2)について質問です。 BCが直径であり、Qが動くことはわかりますが、問題文では「Bから線分PCにひいた垂線のPCとの交点をQ」といってるのでQはPC上にあることになると思うんですけど、なぜ弧BD上を動くことになるんですか? BではなくPC上からではないんですか?

OAD =180*一 CAD 180?ー有GBD =の0BC 芝| G) AG は折財の ・ DB は円周の -和合 だから. AOC=836 Mie )こ a60eメーーニ72 ( 360 10 367. ZDOB = 360 10 72 \ 1 1 1 クワ っ2 い( 3 18。 ZDAB=テ^/2 0 DEB は AAEB の外角だから, ーー ABC十有ZDAB 18*十36 54 [| Q) <zBAC= ZBDC(= 909ょり, 4点A, BC D は 1つの円周上にある。 AB の円周角だから, ACBニンADB あー1308 (2) つねにン2BQC=90'だから, Qは線分BCを 直径とする円周上を動く。 EがAのときQはD に重なり .PがBに近づくとQもBに近づくので, PがBのときQはBに重なると考えてよい。 よ って, Q は BD の上を動く。 BC を直径とする円の中心(BC の中点)を O とす ると. BD の四心角は BOD で, BCD はその 円周角だから ノBOD三22ZBCD=2X60* =120* BC 2 cm より, 円の半径は 1 cm だから, BD の 長きは 27X1X 辻 =今z(cm) 10| (1) 接線と弦のつくる角の定理より, ンPAB = ZC=58" PBA = ZC=58* ノァテー180?一(58?十587) =64" 円の中心を O とする と, 還 丁症 ※はさく へ たか

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英語 高校生

答えを比べたいので、至急どなたか回答してください。回答に自信がないです。

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英語 高校生

英語の長文の回答お願いします。

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英語 高校生

分かる所のみでも良いので、教えてください。お願いします。

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