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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