Sabiendo que el ángulo B es doble que A, en el triángulo ABC, demuestra que AI = b - a

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AI = (p-a)/cos B = (b+c -a)/((b+c)/a) = Ver ejercicios anteriores = (b+c -a)/(a/b) = (b2+ bc - ab) / a = (a2 - ab) / a = a-b

Pepe Martínez, 14/10/2005, Creado con GeoGebra