6.「2026福建福州期中，★☆」如图,将矩形ABCD绕着点C按顺时针方向旋转得到矩形FECG,点B与点E对应,点E恰好落在AD边上,BH⊥CE于点H.
(1)求证:AB=BH.
(2)连接BG交CH于点O,若AB=5,BC=13,求BO的长.

6.解析 (1)证明:$\because$ 四边形 ABCD 是矩形,
$\therefore AD // BC,∠D = 90°,AB = CD,\therefore ∠DEC = ∠BCH,$
$\because BH ⊥ CE,\therefore ∠BHC = 90° = ∠D,$
由旋转的性质得 $CE = CB,\therefore △ EDC ≌ △ CHB(AAS),$
$\therefore BH = CD,\because AB = CD,\therefore AB = BH.$
(2)由旋转的性质得 $∠ECG = ∠BCD = 90°,CD = CG,\therefore BH = CG$.在 $△ HBO$ 和 $△ CGO$ 中, $\begin{cases} ∠OHB = ∠OCG = 90°, \\ ∠HOB = ∠COG, \\ BH = CG, \end{cases}$
$\therefore △ HBO ≌ △ CGO(AAS),\therefore OH = OC,$
在 $\mathrm{Rt}△ BCH$ 中,$BH = 5,BC = 13,$
由勾股定理得 $CH = \sqrt{BC^2-BH^2} = \sqrt{13^2-5^2} = 12,$
$\therefore OH = \frac{1}{2}CH = 6,$
在 $\mathrm{Rt }△ OHB$ 中, 由勾股定理得 $BO = \sqrt{BH^2+OH^2} = \sqrt{5^2+6^2} = \sqrt{61}.$

7.「2026江苏扬州邗江期中改编，★★☆」如图，P是等边三角形ABC内一点，连接PA，PB，PC，将线段BP绕点B顺时针旋转60°得到BQ，连接CQ。
(1)求证:AP=CQ。
(2)若∠APB=150°，PA=5，PB=12，求PC的长。

7.解析 (1)证明:$\because △ ABC$ 为等边三角形,
$\therefore AB = BC,∠ABC = 60°,\therefore ∠ABP+∠CBP = 60°.$
由旋转的性质得 $∠PBQ = 60°,BP = BQ,$
$\therefore ∠CBQ+∠CBP = 60°,\therefore ∠ABP = ∠CBQ.$
$\therefore △ ABP ≌ △ CBQ(SAS),\therefore AP = CQ.$
(2)连接 PQ(图略),
$\because BP = BQ,∠PBQ = 60°,\therefore △ BPQ$ 为等边三角形,
$\therefore PQ = PB = 12,∠BQP = 60°.$
由(1)知 $△ ABP ≌ △ CBQ,$
$\therefore CQ = PA = 5,∠BQC = ∠APB = 150°,$
$\therefore ∠PQC = ∠BQC-∠BQP = 90°,$
$\therefore PC = \sqrt{PQ^2+CQ^2} = \sqrt{12^2+5^2} = 13.$