In the given figure, a circle inscribed in a triangle ABC touches the sides AB, BC and CA at points D, E and F respectively. If AB = 14 cm, BC = 8 cm and CA = 12 cm. Find the lengths AD, BE and CF.

Let AD = x cm, BE = y cm and CF = z cm


As we know that,


Tangents from an external point to a circle are equal,


In given Figure we have


AD = AF = x


[Tangents from point A]


BD = BE = y


[Tangents from point B]6CF = CE = z [Tangents from point C]


Now, Given: AB = 14 cm


AD + BD = 14


x + y = 14


y = 14 - x … [1]


and BC = 8 cm


BE + EC = 8


y + z = 8


14 - x + z = 8 … [From 1]


z = x - 6 [2]


and


CA = 12 cm


AF + CF = 12


x + z = 12 [From 2]


x + x - 6 = 12


2x = 18


x = 9 cm


Putting value of x in [1] and [2]


y = 14 - 9 = 5 cm


z = 9 - 6 = 3 cm


So, we have AD = 9 cm, BE = 5 cm and CF = 3 cm


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