In Fig., a circle inscribed in triangle ABC touches its sides AB,BC and AC at points D,E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, then find the lengths of AD, BE and CF.

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Solution

Given that AB = 12 cm, BC = 8 cm and AC = 10 cm.
Let, AD = AF = p cm, BD = BE = q cm and CE = CF = r cm
(Tangents drawn from an external point to the circle are equal in length)
2(p+ q +r) = AB+BC+AC = AD+DB+BE+EC+AF+FC = 30 cm
p +q +r = 15
AB = AD + DB = p + q = 12 cm
Therefore, r = CF = 15 - 12 = 3 cm.
AC = AF + FC = p + r = 10 cm
Therefore, q = BE = 15 - 10 = 5 cm.
Therefore, p = AD = p + q + r - r - q = 15 - 3 - 5 = 7 cm.

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