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Question
In Fig. 4, 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.
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)
`rArr2(p+q+r)`= AB + BC + AC = AD + DB + BE + EC + AF + FC = 30 cm
`rArr 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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