《2026年课堂作业武汉出版社九年级数学下册人教版》

1.(南京中考)如图，在$\triangle ABC$和$\triangle A'B'C'$中，$D$，$D'$分别是$AB$，$A'B'$上一点，$\frac{AD}{AB}=\frac{A'D'}{A'B'}$.

(1)当$\frac{CD}{C'D'}=\frac{AC}{A'C'}=\frac{AB}{A'B'}$时，求证$\triangle ABC \backsim \triangle A'B'C'$.
(2)当$\frac{CD}{C'D'}=\frac{AC}{A'C'}=\frac{BC}{B'C'}$时，求证$\triangle ABC \backsim \triangle A'B'C'$.

答案： 1.
(1)由$\frac{AD}{AB}=\frac{A'D'}{A'B'}$，得$\frac{AB}{A'B'}=\frac{AD}{A'D'}$，再结合条件$\frac{CD}{C'D'}=\frac{AC}{A'C'}=\frac{AB}{A'B'}$，推出$\frac{CD}{C'D'}=\frac{AC}{A'C'}=\frac{AD}{A'D'}$.利用三边对应成比例证明$\triangle ADC \backsim \triangle A'D'C'$，从而推出$\angle A=\angle A'$.再利用两边对应成比例及其夹角相等，推出$\triangle ABC \backsim \triangle A'B'C'$.
(2)分别过点$D$和$D'$作$DE // BC$交$AC$于点$E$，$D'E' // B'C'$交$A'C'$于点$E'$.$\therefore \triangle ADE \backsim \triangle ABC$.$\therefore \frac{AD}{AB}=\frac{DE}{BC}=\frac{AE}{AC}$.同理$\frac{A'D'}{A'B'}=\frac{D'E'}{B'C'}=\frac{A'E'}{A'C'}$.又$\because \frac{AD}{AB}=\frac{A'D'}{A'B'}$，$\therefore \frac{DE}{BC}=\frac{D'E'}{B'C'}$，$\therefore \frac{DE}{D'E'}=\frac{BC}{B'C'}$.同理$\frac{AE}{AC}=\frac{A'E'}{A'C'}$.$\therefore \frac{AC - AE}{AC}=\frac{A'C'-A'E'}{A'C'}$，$\therefore \frac{EC}{AC}=\frac{E'C'}{A'C'}$，$\therefore \frac{EC}{E'C'}=\frac{AC}{A'C'}$.又$\because \frac{CD}{C'D'}=\frac{AC}{A'C'}=\frac{BC}{B'C'}$，$\therefore \frac{CD}{C'D'}=\frac{EC}{E'C'}=\frac{DE}{D'E'}$.$\therefore \triangle DCE \backsim \triangle D'C'E'$，$\therefore \angle CED=\angle C'E'D'$.又$\because DE // BC$，$\therefore \angle CED+\angle ACB=180^{\circ}$.同理$\angle C'E'D'+\angle A'C'B'=180^{\circ}$.$\therefore \angle ACB=\angle A'C'B'$.又$\because \frac{AC}{A'C'}=\frac{BC}{B'C'}$，$\therefore \triangle ABC \backsim \triangle A'B'C'$.