# 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

[Tangents from point A]

BD = BE = y

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

Now, Given: AB = 14 cm

x + y = 14

y = 14 - x … 

and BC = 8 cm

BE + EC = 8

y + z = 8

14 - x + z = 8 … [From 1]

z = x - 6 

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  and 

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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