Answer is B. 2AD = AB + BC + CA
Property: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.
By the above property,
AE = AD (tangent from A)
AB = AC (tangent from A)
CD = CF (tangent from C)
BF = BE (tangent from B)
Now adding the above equations,
AB + BC + CA = AB + BF + FC + CA
⇒ AB + BC + CA = AB + BE + CD + CA
⇒ AB + BC + CA = AE + AD [∵ AE = AB + BE and AD = AC + CD]
⇒ AB + BC + CA = AD + AD [∵ AD = AE]
⇒ AB + BC + CA = 2AD
Hence, 2AD = AB + BC + CA