已知:△ABC∽△A1B1C1,△A1B1C1∽△A2B2C2.求证:△ABC∽△A2B2C2
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因为△ABC∽△A1B1C1,所以,AB/A1B1=AC/A1C1=BC/B1C1=K1,

所以,AB=K1A1C1,AC=K1A1C1,BC=K1B1C1;

因为△A1B1C1∽△A2B2C2,所以,A1B1/A2B2=A1C1/A2C2=B1C1/B2C2=K2,

所以,A1B1=K2A2B2,A1C1=K2A2C2,B1C1=K2B2C2.

所以,AB=K1K2A2B2,AC=K1K2A2C2,BC=K1K2B2C2,

所以,AB/A2B2=AC/A2C2=BC/B2C2=K1K2,

所以,△ABC∽△A2B2C2.

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