A triangle ABC is drawn to …

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 + 2x
⇒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 (4x + 24 + 32) = 56 + 4x ... (ii)

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

√(14 + x) 48 x = 56 + 4x

Squaring both sides,

48x (14 + x) = (56 + 4x)²

⇒ 48x = [4(14 + x)]²/(14 + x)

⇒ 48x = 16 (14 + x)

⇒ 48x = 224 + 16x

⇒ 32x = 224

⇒ x = 7 cm

Hence, AB = x + 8 = 7 + 8 = 15 cm
CA = 6 + x = 6 + 7 = 13 cm