All the sides BC, CA and AB of a ΔABC touch a circle at D, E and F respectively. AF + BD + CE is equal to:

1
\(\frac{1}{2}\) (Perimeter of ΔABC)
2
\(\frac{1}{3}\) (Perimeter of ΔABC)
3
\(\frac{1}{4}\) (Perimeter of ΔABC)
4
(Perimeter of ΔABC)

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