Given, CD = 6 cm, BD = 8 cm and radius = 4 cm

Join OC, OA and OB.
Let the circle touches the other sides AB and AC at points E and F, respectively.
We know that tangents drawn from an external point to the circle are equal in length.
\(\therefore\) CD = CF = 6 cm            [\(\because\) C is an external point]
BD = BE = 8 cm                [\(\because\) B is an external point]
and AF = AE = x cm (say  [\(\because\) A is an external point]
Area of \(\Delta\)OCB, \(A_1=\frac{1}{2} \times \text { Base } \times \text { Height }\)
\(\begin{aligned} & =\frac{1}{2} \times C B \times O D \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{2} \times 14 \times 4=28 \mathrm{~cm}^2 \\ \end{aligned}\)
\(\begin{aligned} & \quad[\because C B=C D+B D=6+8=14] \end{aligned}\)
Area of \(\Delta\)OCA,
\(\begin{aligned} A_2 & =\frac{1}{2} \times A C \times O F \end{aligned}\)
\(\begin{aligned} =\frac{1}{2}(6+x) \times 4=(12+2 x) \mathrm{cm}^2 \end{aligned}\)
and area of \(\Delta\)OBA,
\(A_3=\frac{1}{2} \times A B \times O E=\frac{1}{2}(8+x) \times 4=(16+2 x) \mathrm{cm}^2\)
Thus, area of  \(\Delta\)ABC
= A₁ + A₂ + A₃ = [28 + (12 + 2x) + (16+ 2x)]
= (56 + 4x) cm²     ...(i)
Now, semi-perimeter of \(\Delta\)ABC=\(\frac{1}{2}\)(AB + BC + CA)
\(\Rightarrow \quad s=\frac{1}{2}(x+8+14+6+x)\)
\(\Rightarrow\) s = (14 + x) cm
Using Heron's formula,
area of \(\Delta\)ABC = \(\begin{aligned} & =\sqrt{s(s-a)(s-b)(s-c)} \end{aligned}\)
\(\begin{aligned} =\sqrt{(14+x)(14+x-14)(14+x-x-6)(14+x-x-8)} \end{aligned}\)
\(\begin{aligned} & =\sqrt{(14+x) \times x \times 8 \times 6} \end{aligned}\)
\(\begin{aligned} =\sqrt{(14+x) 48 x} \end{aligned}\)     ...(ii)
From Eqs. (i) and (ii), we get
\(\sqrt{(14+x) 48 x}=56+4 x=4(14+x)\)
On squaring both sides, we get
(14 + x) 48 x = 4² (14 + x)²
\(\Rightarrow\) 3x = 14 + x
\(\Rightarrow\) 2x = 14
\(\Rightarrow\) x = 7
\(\therefore\)  Length of AC = 6 + x = 6 + 7 = 13 cm
and length of AB = 8 + x = 8 + 7 = 15 cm

\(x={ 58 }^{ \circ }\)

20 cm

Let O and O' be the centres of two circles, tangents AP and AQ intersect these circles in A and the circle with centre O in P and the circle with centre O' in Q. Join PBQ and AB. 

Now, AP is a tangent to the circle with centre O' and AB is the chord.
ஃ  BAP = AQB       .......(i)
[ s in the corresponding alternate segment]
 Again, AQ is a tangent to the circle with centre O and AB is the chord.
ஃ \(\angle \)BAQ = \(\angle \)BPA     .....(ii)
Now, by angle sum property, we have
\(\angle \)BPA + \(\angle \)BAP + \(\angle \)ABP = 180°
\(\angle \)AQB + \(\angle \)BAQ + \(\angle \)QBA = 180°
FRom (iii) and (iv), we have
\(\angle \)BPA + \(\angle \)BPA + \(\angle \)ABP = \(\angle \)AQB + \(\angle \)BAQ + \(\angle \)QBA 
⇒ \(\angle \)ABP = \(\angle \)QBA    [using (i) & (ii)]
hence, AB is the bisector of \(\angle \)PBQ.

Since AB and AC are two tangents from Point A to the two circle.
ஃ  OA ⊥ AC and O'A ⊥ AB
[∵  radius through the point of contact is perpendicular to the tangents]
Let OP intersects AB at M, therefore OM ⊥ AB 
ஃ AM = BM
[∵ ⊥ from the centre of a circle to a chord bisects the chord]
ஃ OM and hence OP is the perpendicular bisector of AB.
Similarly PO' is perpendicular of AC.
Now, in \(\Delta \)ABC
OP is the perpendicular bisector of side AB.
ஃ PA = PB
[∵ any point on the perpendicular bisector is equidistant from the fixed point]
Similarly,  PA = PC
Hence,  PA = PB = PC
ஃ P is equidistant from the three vertices of \(\Delta \)ABC.
Hence, P is the circumcentre of \(\Delta \)ABC.

As, AOPO' is a parallelogram.
ஃ AO || O'P   [opposite sides of a || gm]
ஃ OP ⊥ AB
[ ∵ AO' || OP and O'A ⊥ AB, ஃ OP ⊥ AB]