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$\triangle ABC \backsim \triangle A'B'C'$.

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$\because \frac{AC}{A'C'}=\frac{BC}{B'C'}$，$\therefore \triangle ABC \backsim \triangle A'B'C'$.

(1)证明：
∵$\frac{AD}{AB}=\frac{A'D'}{A'B'}$，
∴$\frac{AB}{A'B'}=\frac{AD}{A'D'}$．
又
∵$\frac{CD}{C'D'}=\frac{AC}{A'C'}=\frac{AB}{A'B'}$，
∴$\frac{CD}{C'D'}=\frac{AC}{A'C'}=\frac{AD}{A'D'}$．
∴$\triangle ADC \backsim \triangle A'D'C'$（三边对应成比例的两个三角形相似）．
∴$\angle A=\angle A'$．
∵$\frac{AC}{A'C'}=\frac{AB}{A'B'}$且$\angle A=\angle A'$，
∴$\triangle ABC \backsim \triangle A'B'C'$（两边对应成比例且夹角相等的两个三角形相似）．
(2)证明：
分别过点$D$和$D'$作$DE // BC$交$AC$于点$E$，$D'E' // B'C'$交$A'C'$于点$E'$．
∵$DE // BC$，
∴$\triangle ADE \backsim \triangle ABC$．
∴$\frac{AD}{AB}=\frac{DE}{BC}=\frac{AE}{AC}$．
同理，$\frac{A'D'}{A'B'}=\frac{D'E'}{B'C'}=\frac{A'E'}{A'C'}$．
∵$\frac{AD}{AB}=\frac{A'D'}{A'B'}$，
∴$\frac{DE}{BC}=\frac{D'E'}{B'C'}$，$\frac{AE}{AC}=\frac{A'E'}{A'C'}$．
∴$\frac{DE}{D'E'}=\frac{BC}{B'C'}$，$\frac{AC-AE}{AC}=\frac{A'C'-A'E'}{A'C'}$，即$\frac{EC}{AC}=\frac{E'C'}{A'C'}$．
∴$\frac{EC}{E'C'}=\frac{AC}{A'C'}$．
∵$\frac{CD}{C'D'}=\frac{AC}{A'C'}=\frac{BC}{B'C'}$，
∴$\frac{CD}{C'D'}=\frac{EC}{E'C'}=\frac{DE}{D'E'}$．
∴$\triangle DCE \backsim \triangle D'C'E'$（三边对应成比例的两个三角形相似）．
∴$\angle CED=\angle C'E'D'$．
∵$DE // BC$，
∴$\angle CED+\angle ACB=180°$．
同理，$\angle C'E'D'+\angle A'C'B'=180°$．
∴$\angle ACB=\angle A'C'B'$．
∵$\frac{AC}{A'C'}=\frac{BC}{B'C'}$且$\angle ACB=\angle A'C'B'$，
∴$\triangle ABC \backsim \triangle A'B'C'$（两边对应成比例且夹角相等的两个三角形相似）．