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.
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.