[i]Comprobar que la composición de dos homotecias de centros diferentes, es una homotecia cuyo centro está alineado con los anteriores y su razón de homotecia es el producto de las razones.[/i][br][br]Dibujamos un triángulo ABC, dos puntos O y O' que utilizaremos como centros de las sucesivas  homotecias, cuyas razones sean 2 y 1'5, respectivamente.[br]Realizamos la primera homotecia del triángulo con centro en O y razón 2 y, al triángulo obtenido le aplicamos una nueva homotecia de razón 1'5 y centro O'. En ambos casos, utilizamos la herramienta [b]Homotecia[/b] [img width=24,height=24]data:image/png;base64,R0lGODlhGAAYAHcAMSH+GlNvZnR3YXJlOiBNaWNyb3NvZnQgT2ZmaWNlACH5BAEAAAAALAIAAwAUABIAhQAAAAAAAAAAGQAAHQAAFQAANgAAOAAAOgAAMwAANwAALgAAJAAAOQAASAAASQAATQAAeQAAawAAgwAAkQAAhwAAkgAAyQAA/wAA9DoAADcAADUAADMAADgAAD0AADEAAEIAAFMAAEcAAFVVVWZmZmdnZ25ubmRkZGVlZWpqatkAAMAAAP8AAPoAAAECAwECAwECAwECAwECAwECAwECAwECAwECAwECAwECAwECAwECAwECAwECAwECAwECAwECAwZBQIBwSCwajaWjEKRsAlSskLJEMnJWLdFU2fE0U86weEwuEwdmoWRiMB8ul0YaQimQFUPC2IGppCMXFmkADwyDg0EAOw==[/img].[br][br][img 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determinar el centro de la composición de las dos homotecias, trazamos dos rectas que unan puntos homotéticos del primer y del tercer triángulo; el punto de intersección O'' corresponde al centro de la[br]homotecia composición.[br][br][img 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RRQAUUVj+LJ5bbwtqM0ErxSpCSroxVlPsRQBqySxRbfMkRNx2ruYDJ9BVf+0YXNwsKyzSW/30RDyfQE4BP41V0OGK40bT554klmSIFZHUMwJ64JrVoAqebfSLA0dskQY5lWZ/mQe23IJ/GlNrPI04lvHMcowixqEMY9mHOferVFAFVdOtAsIeESmD/VvN+8ZT65bJz71aoooAKKKKACiiigAqOaaO3geeZxHHGpZ2Y4AA6mpKx/EXzR6dGeUkv4VdT0YZJwR3GQPyoAfp8Ml9c/2tdoUyCLSFhzEh/iI/vN+gwPWtWiigAooooAKKKKAKl5pttfFXkVkmT7k8Z2yJ9CP5dKrfab/TeL1DeW4/5eIU+dR/toOv1X8hWpRQBFb3MF3CJreVJY26MhyKlrnLn/AEfx3YJB+6W4t5WmCfKJCOhbHUj3ro6ACiiigAooooAKKKKACiiigAooooA//9k=[/img][br][br]Para hallar la razón k de la homotecia composición, bastará con aplicar la definición de homotecia en la  que se cumple la relación: [br][br][img width=69,height=41]data:image/png;base64,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[/img][br][br]Podemos comprobar que el resultado del cociente anterior es 3 que corresponde con el valor del producto de las dos homotecias 2 x 1,5 = 3.[br]El centro de la homotecia composición es el punto O’’ intersección de las dos rectas dibujadas entre los puntos AA’’ y CC’’.[br][br]