$解:如图,过点C作CG//AB,过点D作DH//AB,\ $ $则CG//DH,∠BCG=∠B= 25°.$ $又∠BCD=45°,\ $ $∴∠DCG=45°-25°=20°.\ $ $∵CG//DH,$ $∴∠CDH=∠DCG=20°,\ $ $∴∠HDE=∠CDE-∠CDH =30°-20°=10°,$ $∴∠E=∠HDE=10°,\ $ $∴HD//EF,$ $∴AB//EF.$
$证明:根据平移性质可知A'C'//AC,AA'//CC',\ $ $∴∠A'=∠BAC,∠BAC=∠ACC'.\ $ $∴∠A'=∠ACC'.\ $ $∵∠ACC'+∠CAC'+∠AC'C=180°,\ $ $∴∠A'+∠CAC'+∠AC'C=180°.$
$解:∠CAC'=x+y.证明如下:$ $如图,过点A作AD//BC,交CC'于点D.\ $ $根据平移性质可知B'C'//BC,$ $∴B'C'//AD//BC, ∴∠AC'B'=∠C'AD,∠ACB=∠DAC. ∴∠CAC‘=∠C'AD+∠CAD=∠AC'B'+∠ACB=x+y,$ $即∠CAC'=x+y.\ $
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