$解：?(1)?如圖所示$
$?(2)∠A=∠A，??△ABC∽△A'B'C'?$
$在?AB?上截取?AB''= A'B'，?過點(diǎn)?B''?作?B''C''//BC，?交?AC?于點(diǎn)?C''?$
$在?△ABC?和?△AB''C''?中$
$∵?B''C''//BC?$
$∴?△ABC \sim △AB''C''?$
$∴?\frac {AB}{AB''}=\frac {BC}{B''C''}=\frac {CA}{C''A}?$
$∵?\frac {AB}{A'B'}=\frac {BC}{B'C'}=\frac {CA}{C'A'}，??AB''= A'B'?$
$∴?B''C''= B'C'，??C''A= C'A'，??△AB''C''≌△A'B'C'?$
$∴?△ABC \sim △A'B'C'?$
$?(3)?假設(shè)?AB＞A'B'，?在?AB?上截取?AB''= A'B'，?過點(diǎn)?B''?作?B''C''//BC，?交?AC?于點(diǎn)?C''?$
$在?△ABC?和?△AB''C''?中$
$∵?B''C''//BC?$
$∴?△ABC \sim △AB''C''?$
$∴?\frac {AB}{AB''}=\frac {BC}{B''C''}=\frac {CA}{C''A}?$
$∵?\frac {AB}{A'B'}=\frac {BC}{B'C'}=\frac {CA}{C'A'}，??AB''= A'B'?$
$∴?B''C''= B'C'，??C''A= C'A'，??△AB''C''≌△A'B'C'?$
$∴?△ABC \sim △A'B'C'?$
