It is given that a circle inscribed in a ΔABC having sides.
AB = 8 cm,
BC = 10 cm
and AC = 12 cm
Since, tangents drawn from an external points are equal.
So, AD = AF
BD = BE
and CE = CF
Let AD = AF = x
BD = BE = y
and CE = CF = z
AB = 8
⇒ AD + BD = 8
⇒ x + y = 8 ......(i)
∵ BC = 10
⇒ BE + CE = 10
⇒ y + z = 10 ....(ii)
∵ AC = 12
⇒ AF + CF = 12
⇒ x + z = 12 ...(iii)
Adding (i), (ii) and (iii), we get
2x + 2y + 2z = 30
⇒ 2(x + y + z) = 30
⇒ x + y + z = 15 ...(iv)
Putting the value of (i) in (iv), we get
8 + z = 15
⇒ z = 15 – 8
= 7 cm.
Putting the value of (ii) in (iv)
x + 10 = 15
⇒ x = 5 cm
Putting the value of (iii) in (iv), we get
y + 12 = 15 y = 3 cm
Hence, x = AD = 5 cm
y = BE = 3 cm and z = CF = 7 cm