Sublimation
Explanation:

Given, in â–³ABC, circle touch the triangle at point D, F and E respectively and let the lengths of the segment AF be x.
So, BF = BD = 6Â cm [Tangent drawn from same point are equal]
CE = CD = 9Â cm [Tangent drawn from same point are equal ]
and AE = AF = x cm[ Tangent drawn from same point are equal]
Now, Area of △OBC = $ \displaystyle \dfrac{1}{2}$ ​× BC × OD
​= $ \displaystyle \dfrac{1}{2}$​ × (6 + 9) × 3
= $ \displaystyle \dfrac{45}{2}$​$ \displaystyle \text{c}{{\text{m}}^{2}}$​
Area of △OCA = $ \displaystyle \dfrac{1}{2}$​ × AC × OE
​= $ \displaystyle \dfrac{1}{2}$​ × (9 + x) × 3
= $ \displaystyle \dfrac{3}{2}$​(9 + x)$ \displaystyle \text{c}{{\text{m}}^{2}}$​
Area of △BOA = $ \displaystyle \dfrac{1}{2}$ ​× AB × OF
​= $ \displaystyle \dfrac{1}{2}$ ​× (6 + x) × 3
= $ \displaystyle \dfrac{3}{2}$(6 + x)$ \displaystyle \text{c}{{\text{m}}^{2}}$​
Area of â–³ABC = 54 $ \displaystyle \text{c}{{\text{m}}^{2}}$Â [Given]
∵ Area of △ABC = Area of △OBC + Area of △OCA + Area of △BOA.
54 = $ \displaystyle \dfrac{45}{2}$​ + $ \displaystyle \dfrac{3}{2}$​(9 + x) + $ \displaystyle \dfrac{3}{2}$(6 + x)
⇒ 54 × 2 = 45 + 27 + 3x + 18 + 3x
⇒ 108 − 45 − 27 − 18 = 6x
⇒ 6x = 18
⇒ x = 3
So, AB = AF + FB = x + 6 = 3 + 6 = 9Â cm
and AC = AE + EC = x + 9 = 3 + 9 = 12Â cm
Hence, lengths of AB and AC are 9Â cm and 12Â cm respectively.