《2026年课堂作业武汉出版社九年级数学下册人教版》

23. 如图，在$ Rt \triangle ABC$中，$\angle ACB = 90°$，$AC = BC$，$P$为$\triangle ABC$内部一点，且$\angle APB = \angle BPC = 135°$.
(1) 求证$\triangle PAB \backsim \triangle PBC$.
(2) 求证$PA = 2PC$.
(3) 若点$P$到三角形的边$AB$，$BC$，$CA$的距离分别为$h_1$，$h_2$，$h_3$，求证$h_1^2 = h_2h_3$.

答案：
23.
(1)推出∠PBC = ∠PAB,结合∠APB = ∠BPC = 135°得△PAB∽△PBC.
(2)由
(1)得$\frac{PA}{PB}$ = $\frac{PB}{PC}$ = $\frac{AB}{BC}$,
则PB = $\sqrt{2}$PC,PA = $\sqrt{2}$PB,
∴PA = 2PC.
(3)如图,过点P作PF⊥AB,PD⊥BC,PE⊥AC,垂足分别为F,D,E,则PF = h₁,PD = h₂,PE = h₃.
由∠APC = ∠ACB = 90°,
推出Rt△AEP∽Rt△CDP,
∴$\frac{PE}{DP}$ = $\frac{AP}{PC}$ = 2,即$\frac{h₃}{h₂}$ = 2.
∴h₃ = 2h₂.
∵△PAB∽△PBC,
∴$\frac{h₁}{h₂}$ = $\frac{AB}{BC}$ = $\sqrt{2}$,即h₁ = $\sqrt{2}$h₂.
∴h₁² = 2h₂² = 2h₂h₂ = h₂h₃.

24. (1)【问题背景】如图①，在矩形$ABCD$中，点$E$，$F$分别是$AB$，$BC$的中点，连接$BD$，$EF$，求证$\triangle BCD \backsim \triangle FBE$.
(2)【问题探究】如图②，在四边形$ABCD$中，$AD // BC$，$\angle BCD = 90°$，点$E$是$AB$的中点，点$F$在边$BC$上，$AD = 2CF$，$EF$与$BD$交于点$G$，求证$BG = FG$.
(3)【问题拓展】如图③，在“问题探究”的条件下，连接$AG$，$AD = CD$，$AG = FG$，直接写出$\frac{EG}{GF}$的值.

答案：
24.
(1)如图①所示,

∵四边形ABCD是矩形,
∴∠BCD = ∠FBE = 90°,AB = CD.
∵点E,F分别是AB,BC的中点,
∴AB = 2BE,BC = 2FB.
∴$\frac{BC}{FB}$ = $\frac{CD}{BE}$.
∴△BCD∽△FBE.
(2)如图②所示,分别过点A,E作BC的垂线,垂足分别为I,H,则EH//AI.

∴△BEH∽△BAI.
∴$\frac{BE}{BA}$ = $\frac{BH}{BI}$ = $\frac{EH}{AI}$.
∵E是AB的中点,
∴HI = $\frac{1}{2}$BI,EH = $\frac{1}{2}$AI.
∵AD//BC,∠BCD = 90°,
∴四边形AICD是矩形.
∴AI = CD,AD = IC.
∵AD = 2CF,
∴IF = $\frac{1}{2}$IC.
∴HF = $\frac{1}{2}$BC.
∴$\frac{FH}{BC}$ = $\frac{EH}{DC}$.
又
∵∠EHF = ∠DCB,
∴△EHF∽△DCB.
∴∠EFB = ∠DBC.
∴BG = FG.
(3)$\frac{\sqrt{5}}{5}$.
提示:如图③所示,过点F作FM⊥AD于点M,则四边形MFCD是矩形,连接AF.

∵AD = 2CF = CD,
∴AM = MD = FC = $\frac{1}{2}$AD.
设AD = 2a,则MF = CD = 2a,AM = a.
在Rt△AMF中,AF = $\sqrt{a²+(2a)²}$ = $\sqrt{5}$a.
∵AG = FG,由
(2)知BG = FG,
∴AG = BG.
又
∵E是AB的中点,
∴EF垂直平分AB.
∴AF = BF,∠BEG = 90°.
又可证△AFG≌△BFG(SSS).
设∠GBF = ∠GFB = α,则∠GAF = ∠GFA = α.
∴∠BGE = ∠GBF + ∠GFB = 2α.
又
∵AD//BC,
∴∠MAF = ∠AFB = ∠GFA + ∠GFB = 2α.
∴∠MAF = ∠EGB.
又
∵∠BEG = ∠AMF = 90°,
∴△BEG∽△FMA.
∴$\frac{EG}{GF}$ = $\frac{EG}{BG}$ = $\frac{AM}{AF}$ = $\frac{a}{\sqrt{5}a}$ = $\frac{\sqrt{5}}{5}$.