典例5(2024·常州溧阳一模)如图,在Rt△ABC中,∠ACB = 90°,BC = 2,点D在AC上,连接BD,使得BD = AC,以AC为边向外作△ACE.若CE//BD,tanE = 2,则边AE的长为________.

如图，过点A作$AM\perp CE$于点M. $\because AM\perp CE$，
$\therefore \angle AMC = 90^{\circ}$. $\because \angle ACB = 90^{\circ}$，$\therefore \angle ACB=\angle AMC$.
$\because CE// BD$，$\therefore \angle BDC=\angle ACM$. 在$\triangle BCD$和$\triangle AMC$中，
$\begin{cases} \angle BCD=\angle AMC = 90^{\circ}, \\ \angle BDC=\angle ACM, \\ BD = AC, \end{cases}$ $\therefore \triangle BCD\cong \triangle AMC(AAS)$. $\therefore BC = AM = 2$. 在$Rt\triangle AME$中，$\tan E=\frac{AM}{ME}=2$，$\therefore ME=\frac{2}{2}=1$.
$\therefore AE=\sqrt{ME^{2}+AM^{2}}=\sqrt{1^{2}+2^{2}}=\sqrt{5}$.

典例6(2024·苏州)如图①是某种可调节支撑架,BC为水平固定杆，竖直固定杆AB⊥BC,活动杆AD可绕点A旋转，CD为液压可伸缩支撑杆,已知AB = 10cm,BC = 20cm,AD = 50cm.
(1)如图②，当活动杆AD处于水平状态时,求可伸缩支撑杆CD的长度(结果保留根号);
(2)如图③,当活动杆AD绕点A由水平状态按逆时针方向旋转角度α,且tanα = $\frac{3}{4}$(α为锐角),求此时可伸缩支撑杆CD的长度(结果保留根号).

(1) 如图①，过点C作$CE\perp AD$，垂足为E. 由题意，得$AB = CE = 10\ cm$，$BC = AE = 20\ cm$. $\because AD = 50\ cm$，$\therefore ED = AD - AE = 50 - 20 = 30(cm)$. 在$Rt\triangle CED$中，$CD=\sqrt{CE^{2}+DE^{2}}=\sqrt{10^{2}+30^{2}}=10\sqrt{10}(cm)$. $\therefore$可伸缩支撑杆CD的长度为$10\sqrt{10}\ cm$. (2) 如图②，过点D作$DF\perp BC$，交BC的延长线于点F，交$AD'$于点G. 由题意，得$AB = FG = 10\ cm$，$AG = BF$，$\angle AGD = 90^{\circ}$. 在$Rt\triangle ADG$中，$\tan \alpha=\frac{DG}{AG}=\frac{3}{4}$，$\therefore$设$DG = 3x\ cm$，则$AG = 4x\ cm$. $\therefore AD=\sqrt{AG^{2}+DG^{2}}=\sqrt{(4x)^{2}+(3x)^{2}}=5x(cm)$. $\because AD = 50\ cm$，$\therefore 5x = 50$，解得$x = 10$. $\therefore AG = 40\ cm$，$DG = 30\ cm$. $\therefore DF = DG + FG = 30 + 10 = 40(cm)$. $\therefore BF = AG = 40\ cm$. $\because BC = 20\ cm$，$\therefore CF = BF - BC = 40 - 20 = 20(cm)$. 在$Rt\triangle CFD$中，$CD=\sqrt{CF^{2}+DF^{2}}=\sqrt{20^{2}+40^{2}}=20\sqrt{5}(cm)$. $\therefore$此时可伸缩支撑杆CD的长度为$20\sqrt{5}\ cm$.

[变式](2024·苏州一模)如图,某学习小组在学习了解直角三角形及其应用的知识后，尝试利用所学知识测量河对岸大树AB的高度,小组同学在点C处测得大树顶端A的仰角为45°,再从点C出发沿斜坡走3$\sqrt{10}$m 到达斜坡上点D,在点D处测得树顶端A 的仰角为30°.若斜坡CF的坡比i = 1:3(点E,C,B在同一水平线上).求:
(1)从点C到点D的过程中上升的高度;
(2)大树AB的高度(结果保留根号).

(1) 如图，过点D作$DG\perp BE$，垂足为G. $\because$斜坡CF的坡比$i = 1:3$，$\therefore \frac{DG}{CG}=\frac{1}{3}$. $\therefore$设$DG = x\ m$，则$CG = 3x\ m$. 在$Rt\triangle DCG$中，$CD=\sqrt{DG^{2}+CG^{2}}=\sqrt{x^{2}+(3x)^{2}}=\sqrt{10}x(m)$.
$\because CD = 3\sqrt{10}\ m$，$\therefore \sqrt{10}x = 3\sqrt{10}$，解得$x = 3$. $\therefore DG = 3\ m$，$CG = 9\ m$. $\therefore$从点C到点D的过程中上升的高度为3 m. (2) 如图，过点D作$DH\perp AB$，垂足为H. 由题意，得$DG = BH = 3\ m$，$DH = BG$. 设$BC = x\ m$. $\because CG = 9\ m$，$\therefore DH = BG = CG + BC = (x + 9)\ m$. 在$Rt\triangle ABC$中，$\angle ACB = 45^{\circ}$. $\therefore AB = BC\cdot \tan 45^{\circ}=x\ m$. 在$Rt\triangle ADH$中，$\angle ADH = 30^{\circ}$，$\therefore AH = DH\cdot \tan 30^{\circ}=\frac{\sqrt{3}}{3}(x + 9)\ m$. $\because AH + BH = AB$，$\therefore \frac{\sqrt{3}}{3}(x + 9)+3 = x$，解得$x = 6\sqrt{3}+9$. $\therefore AB=(6\sqrt{3}+9)m$. $\therefore$大树AB的高度为$(6\sqrt{3}+9)m$.