In ΔABC,

Length of two tangents drawn from the same point to the circle are equal,

∴ CF = CD = 6cm

∴ BE = BD = 8cm

∴ AE = AF = *x*

We observed that,

AB = AE + EB = *x* + 8

BC = BD + DC = 8 + 6 = 14

CA = CF + FA = 6 +* x*

Now semi perimeter of circle s,

⇒ 2s = AB + BC + CA

= *x* + 8 + 14 + 6 + *x*

= 28 + 2*x*

⇒s = 14 + *x*

Area of ΔABC = √s (s - a)(s - b)(s - c)

= √(14 + *x*) (14 + *x *-* *14)(14 + *x *-* x* - 6)(14 + *x *-* x - *8)

= √(14 + *x*) (*x*)(8)(6)

= √(14 + *x*) 48 *x* ... (i)

also, Area of ΔABC = 2×area of (ΔAOF + ΔCOD + ΔDOB)

= 2×[(1/2×OF×AF) + (1/2×CD×OD) + (1/2×DB×OD)]

= 2×1/2 (4*x *+ 24 + 32) = 56 + 4*x *... (ii)

Equating equation (i) and (ii) we get,

√(14 + *x*) 48 *x *= 56 + 4*x*

Squaring both sides,

48*x* (14 + *x*) = (56 + 4*x*)^{2}

⇒ 48*x = *[4(14 + x)]^{2}/(14 + *x*)

⇒ 48*x = *16 (14 + *x*)

⇒ 48*x = *224 + 16*x*

⇒ 32*x = *224

⇒ *x = *7 cm

Hence, AB = *x* + 8 = 7 + 8 = 15 cm

CA = 6 + *x* = 6 + 7 = 13 cm