Given:

AB = 12 cm

BC = 8 cm

AC = 10 cm

Property: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By the above property,

AD = AF (tangent from A)

BD = BE (tangent from B)

CF = CE (tangent from C)

Clearly,

AB = AD + DB = 12 cm

BC = BE + EC = 8 cm

AC = AF + FC = 10 cm

Now,

AB – BC = 12 cm – 8 cm

⇒ (AD + DB) – (BE + EC) = 12 cm – 8 cm

⇒ AD + DB – BE – EC = 12 cm – 8 cm

⇒ AD + BE – BE – CF = 12 cm – 8 cm [∵ DB = BE and CF = CE]

⇒ AD – CF = 12 cm – 8 cm

⇒ AD – (10 cm – AF) = 12 cm – 8 cm [∵AF + FC = 10 cm ⇒ FC = 10 cm – AF]

⇒ AD – (10 cm – AF) = 4 cm

⇒ AD – 10 cm + AF = 4 cm

⇒ AD + AD = 4 cm + 10 cm [∵ AD = AF]

⇒ 2AD = 14 cm

⇒ AD = 7 cm

Hence, AD = 7 cm