A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.

As we know, the lengths of tangents drawn from an external point to a circle are equal, 

So, CF = CD = 6 cm

BE = BD = 8 cm

Let AE = AF = x

we have figure as:

Now, it can be seen that,

AB = EB + AE = 8 + x

CA = CF + FA = 6 + x

BC = DC + BD = 6 + 8 = 14

Now we can calculate the semi-perimeter “s” as:

2s = AB + CA + BC

By substituting the known values, we get,

2s = 28 + 2x

s = 14 + x

As we know, $Area of ∆ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

Substituting the known values and by solving, we get,

$Area of ∆ABC=\sqrt{\left(14+x\right)\left(48x\right)}$............ (i)

Also, the area of △ABC = 2 × area of (△AOF + △COD + △DOB)

= 2×[(½×OF×AF)+(½×CD×OD)+(½×DB×OD)]

= 2×½(4x+24+32) = 56+4x …………..(ii)

Now from (i) and (ii), we get,

$\sqrt{\left(14+x\right)\left(48x\right)}= 56+4x$ 

Now, square both sides,

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

$48x =\frac{ {\left[4\right(14+x\left)\right]}^2}{(14+x)}$

$48x =\left[16\right(14+x\left)\right]$

48x = 224+16x

32x = 224

x = 7 cm

So, AB = 8+x

That is AB = 15 cm

And, CA = x + 6 =13 cm.