Question

17．如圖，以BC為直徑作半圓$\widehat{BAC}$，A為半圓的中點(diǎn)，現(xiàn)將半圓連同直徑繞點(diǎn)A順時(shí)針旋轉(zhuǎn)45°，記點(diǎn)B、C的對(duì)應(yīng)點(diǎn)分別為B′、C′，連接B′C，BC′，則$\frac{B′C}{BC′}$=（　�。�

• A． $\frac{\sqrt{2}}{4}$
• B． $\frac{2}{5}$
• C． $\sqrt{2}-1$
• D． 2$-\sqrt{2}$

Answer 1

分析 連結(jié)AB、AB′、AC、AC′．根據(jù)旋轉(zhuǎn)的性質(zhì)及圓周角定理得出∠BAC=90°，∠BAB′=∠CAC′=∠B′AC=45°，∠BAC′=135°，AB=AB′=AC=AC′．設(shè)半圓的半徑為r．在△B′AC中，利用余弦定理求出B′C²=AB′²+AC²-2AB′•AC•cos∠B′AC=（2-$\sqrt{2}$）r²，在△BAC′中，求出BC′²=AB²+AC′²-2AB•AC′•cos∠BAC′=（2+$\sqrt{2}$）r²，進(jìn)而得出$\frac{B′C}{BC′}$=$\sqrt{2}$-1．

解答 解：如圖，連結(jié)AB、AB′、AC、AC′．
∵將半圓連同直徑繞點(diǎn)A順時(shí)針旋轉(zhuǎn)45°，記點(diǎn)B、C的對(duì)應(yīng)點(diǎn)分別為B′、C′，
∴∠BAB′=∠CAC′=45°，AB=AB′=AC=AC′．
∵BC為半圓$\widehat{BAC}$的直徑，
∴∠BAC=90°，
∴∠B′AC=∠BAC-∠BAB′=90°-45°=45°，∠BAC′=∠BAC+∠CAC′=90°+45°=135°．
設(shè)半圓的半徑為r．
∵在△B′AC中，AB′=AC=r，∠B′AC=45°，
∴B′C²=AB′²+AC²-2AB′•AC•cos∠B′AC=r²+r²-2r•r•$\frac{\sqrt{2}}{2}$=2r²-$\sqrt{2}$r²=（2-$\sqrt{2}$）r²，
∵在△BAC′中，AB=AC′=r，∠BAC′=135°，
∴BC′²=AB²+AC′²-2AB•AC′•cos∠BAC′=r²+r²+2r•r•$\frac{\sqrt{2}}{2}$=2r²+$\sqrt{2}$r²=（2+$\sqrt{2}$）r²，
∴$\frac{B′{C}^{2}}{BC{′}^{2}}$=$\frac{（2-\sqrt{2}）{r}^{2}}{（2+\sqrt{2}）{r}^{2}}$=$\frac{（2-\sqrt{2}）^{2}}{2}$，
∴$\frac{B′C}{BC′}$=$\frac{2-\sqrt{2}}{\sqrt{2}}$=$\sqrt{2}$-1．
故選C．

點(diǎn)評(píng) 本題考查了旋轉(zhuǎn)的性質(zhì)：對(duì)應(yīng)點(diǎn)到旋轉(zhuǎn)中心的距離相等；對(duì)應(yīng)點(diǎn)與旋轉(zhuǎn)中心所連線段的夾角等于旋轉(zhuǎn)角；旋轉(zhuǎn)前、后的圖形全等．也考查了圓周角定理，余弦定理．準(zhǔn)確作出輔助線是解題的關(guān)鍵．