In the given figure, AB, BC and CA are tangents to the given circle. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the lengths of AD, BE and CF.

As, tangents from an exterior fixed point to a circle are equal in length,

Let,

AD = AF = x cm

BD = BE = y cm

CF = CE = z cm

From figure,

⇒ AD + BD = AB

⇒ x + y = 12 ………(1)

⇒ AF + CF = AC

⇒ x + z = 10 ……..(2)

⇒ BE + CE = BC

⇒ y + z = 8 ……….(3)

Subtracting eq (3) from (1), we get :

⇒ x + y - (y + z) = 12 - 8

⇒ x - z = 4 …………(4)

Adding equation (2) and (4), we get :

⇒ (x + z) + (x - z) = 10 + 4

⇒ 2x = 14

⇒ x = $\dfrac{14}{2}$

⇒ x = 7.

Substituting value of x in equation (1), we get :

⇒ 7 + y = 12

⇒ y = 12 - 7

⇒ y = 5.

Substituting value of y in equation (3), we get :

⇒ y + z = 8

⇒ 5 + z = 8

⇒ z = 8 - 5

⇒ z = 3.

∴ AD = x = 7, BE = y = 5 and CF = z = 3.

Hence, AD = 7 cm, BE = 5 cm and CF = 3 cm.