第41页 - 亮点给力提优课时作业本九年级数学江苏版

C

(3,-1)

$\frac{25}{2}$

$\frac{4\pi}{3}$

D

(7,4)或(1,4)或(6,5)

解: 如图，作△ABC的外接圆⊙O，连接OA，OB，OC，延长AO交⊙O于点H，过点O作OE⊥BC于点E，OF⊥AD于点F，则BE = CE = $\frac{1}{2}BC，$∠OED = ∠OFA = ∠OFD = 90°. 因为AD是△ABC的高，所以AD⊥BC，所以∠EDF = 90°，所以四边形OEDF是矩形，所以OE = DF，OF = DE. 因为BD = 3，CD = 1，所以BC = BD + CD = 4，所以BE = CE = 2. 因为OA = OB = OC，所以∠OAB = ∠OBA，∠OAC = ∠OCA，所以∠BOH = ∠OAB + ∠OBA = 2∠OAB，∠COH = ∠OAC + ∠OCA = 2∠OAC，所以∠BOC = ∠BOH + ∠COH = 2(∠OAB + ∠OAC) = 2∠BAC. 因为∠BAC = 45°，所以∠BOC = 90°. 设OA = OB = OC = x，则BC = $\sqrt{OB^{2}+OC^{2}}=\sqrt{2}x，$所以$\sqrt{2}x = 4，$解得x = $2\sqrt{2}，$所以OA = OB = OC = $2\sqrt{2}，$所以DF = OE = $\sqrt{OB^{2}-BE^{2}} = 2，$OF = DE = CE - CD = 1，所以AF = $\sqrt{OA^{2}-OF^{2}}=\sqrt{7}，$所以AD = AF + DF = $\sqrt{7}+2.$

解: (1) △ABC是等腰三角形. 证明如下：过点D作DE⊥AB于点E，DF⊥AC于点F. 因为AD是△ABC的角平分线，所以DE = DF. 因为AD是△ABC的中线，所以BD = CD. 在Rt△BDE和Rt△CDF中，$\begin{cases}BD = CD\\DE = DF\end{cases}，$所以Rt△BDE≌Rt△CDF，所以∠B = ∠C，所以△ABC是等腰三角形.

 (2) 直线AD经过△ABC的外接圆的圆心. 证明如下：因为△ABC是等腰三角形，AD是中线，所以AD⊥BC，所以AD垂直平分BC，所以直线AD经过△ABC的外接圆的圆心.