In Fig. 3, 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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Fig.3


In the fig.3,

AD and AF are making a pair of tangents drawn from vertex A of Δ ABC on the circle.


AD = AF = x [LET]


[ Tangents drawn from an exterior point to the circle are equal in length]


BD and BE are making a pair of tangents drawn from vertex B of Δ ABC on the circle.


BD = BE = y [LET]


[ Tangents drawn from an exterior point to the circle are equal in length]


CE and CF are making a pair of tangents drawn from vertex C of Δ ABC on the circle.


CE = CF = z [LET]


[ Tangents drawn from an exterior point to the circle are equal in length]


From fig.3-


AF+CF = x+z = 10 cm …(1)


BE+CE = y+z = 8 cm …(2)


AD+BD = x+y = 12 cm …(3)


Adding equations (1), (2) and (3), we get-


2(x+y+z) = 30


x+y+z = 15 …(4)


Subtracting equations (1), (2) and (3) from (4) one bye one, we get-


x = 7 cm, y = 5 cm, z = 3 cm


Thus, Length of AD = 7 cm


Length of BE = 5 cm


and, Length of CF = 3 cm.


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