A triangle ABC is drawn to …

Let M and N be the points where AB and AC touch the circle respectively.
Tangents drawn from an external point to a circle are equal
{tex} \Rightarrow{/tex} AM = AN
BD = BM = 8 and DC = NC = 6 cm
In {tex} \Delta{/tex}ABC,
{tex} ar(\Delta ABC) = \frac{1}{2} \times BC \times AD{/tex}
{tex}\Rightarrow 84 = \frac{1}{2} \times 14 \times AD{/tex}
{tex}\Rightarrow \frac{{84 \times 2}}{{14}} = AD{/tex}
{tex} \Rightarrow{/tex} AD = 12 cm
In right {tex} \Delta{/tex}ABD,
AB² = BD² + AD²
{tex} \Rightarrow{/tex} AB² = 8² + 12²
{tex} \Rightarrow{/tex} AB² = 64 + 144
{tex} \Rightarrow{/tex} AB = {tex} \sqrt {208}{/tex}
{tex} \Rightarrow{/tex} AB = 14.42 cm
So, AM = AB - BM = 14.42 - 8 = 6.42 cm
{tex} \Rightarrow{/tex} AN = AM = 6.42 cm
So, AC = AN + NC = 6.42 + 6 = 12.42 cm
Hence, AB = 14.42 cm and AC = 12.42 cm