21. （2024·苏州）如图，在△ABC中，AB = 4$\sqrt{2}$，D为AB的中点，∠BAC = ∠BCD，cos∠ADC = $\frac{\sqrt{2}}{4}$，⊙O是△ACD的外接圆. 求：
（1）BC的长；
（2）⊙O的半径.
　　　　

(1) $\because\angle BAC=\angle BCD$，$\angle B=\angle B$，$\therefore\triangle BAC\sim\triangle BCD$。$\therefore\frac{BC}{BD}=\frac{BA}{BC}$。$\because AB = 4\sqrt{2}$，$D$为$AB$的中点，$\therefore BD = AD = 2\sqrt{2}$。$\therefore\frac{BC}{2\sqrt{2}}=\frac{4\sqrt{2}}{BC}$。$\therefore BC = 4$(负值舍去) (2) 如图，过点$A$作$AE\perp CD$，垂足为$E$，连接$CO$并延长，交$\odot O$于点$F$，连接$AF$。$\because$在$Rt\triangle AED$中，$\cos\angle ADC=\frac{DE}{AD}=\frac{\sqrt{2}}{4}$，$AD = 2\sqrt{2}$，$\therefore DE = 1$。$\therefore$由勾股定理，得$AE=\sqrt{AD^{2}-DE^{2}}=\sqrt{7}$。$\because\triangle BAC\sim\triangle BCD$，$\therefore\frac{AC}{CD}=\frac{AB}{CB}=\frac{4\sqrt{2}}{4}=\sqrt{2}$。设$CD = x$，则$AC=\sqrt{2}x$，$CE = x - 1$。$\because$在$Rt\triangle ACE$中，由勾股定理，得$AC^{2}=CE^{2}+AE^{2}$，$\therefore(\sqrt{2}x)^{2}=(x - 1)^{2}+(\sqrt{7})^{2}$，即$x^{2}+2x - 8 = 0$，解得$x_{1}=2$，$x_{2}=-4$(不合题意，舍去)。$\therefore CD = 2$，$AC = 2\sqrt{2}$。$\because\overset{\frown}{AC}=\overset{\frown}{AC}$，$\therefore\angle AFC=\angle ADC$。$\because CF$为$\odot O$的直径，$\therefore\angle CAF = 90^{\circ}$。$\therefore\sin\angle AFC=\frac{AC}{CF}=\sin\angle ADC=\frac{AE}{AD}$。$\therefore\frac{2\sqrt{2}}{CF}=\frac{\sqrt{7}}{2\sqrt{2}}$，解得$CF=\frac{8\sqrt{7}}{7}$。$\therefore\odot O$的半径为$\frac{4\sqrt{7}}{7}$

22. （2023·泸州）如图，某数学兴趣小组为了测量古树DE的高度，采用的方法如下：先从与古树底端D在同一水平线上的点A出发，沿坡度为i = 2∶$\sqrt{3}$的斜坡AB前进20$\sqrt{7}$ m到达点B，再沿水平方向继续前进一段距离后到达点C. 在点C处测得古树DE的顶端E的俯角为37°，底部D的俯角为60°，求古树DE的高度（结果保留根号，参考数据：sin37°≈$\frac{3}{5}$，cos37°≈$\frac{4}{5}$，tan37°≈$\frac{3}{4}$）.
　　

如图，过点$B$作$BF\perp AD$于点$F$。$\because i = 2:\sqrt{3}$，$\therefore$可设$BF = 2k$m，$AF=\sqrt{3}k$m。$\because$在$Rt\triangle ABF$中，$BF^{2}+AF^{2}=AB^{2}$，$AB = 20\sqrt{7}$m，$\therefore(2k)^{2}+(\sqrt{3}k)^{2}=(20\sqrt{7})^{2}$，解得$k = 20$(负值舍去)。$\therefore BF = 40$m。延长$BC$、$DE$交于点$H$。$\because BC$是水平线，$DE$是铅直线，$\therefore DH\perp CH$，四边形$BFDH$是矩形。$\therefore DH = BF = 40$m。$\because$在$Rt\triangle CDH$中，$\tan\angle DCH=\frac{DH}{CH}$，$\therefore CH=\frac{DH}{\tan\angle DCH}=\frac{40\sqrt{3}}{3}$m。$\because$在$Rt\triangle CEH$中，$\tan\angle ECH=\frac{EH}{CH}$，$\therefore EH = CH\cdot\tan\angle ECH=\frac{40\sqrt{3}}{3}\cdot\tan37^{\circ}\approx\frac{40\sqrt{3}}{3}\times\frac{3}{4}=10\sqrt{3}$(m)。$\therefore DE = DH - EH=(40 - 10\sqrt{3})$m。$\therefore$古树$DE$的高度为$(40 - 10\sqrt{3})$m