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In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at points D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the lengths of 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]
CF = CE = z [Tangents from point C]
Now, Given: AB = 12 cm
AD + BD = 12
x + y = 12
y = 12 – x…. [1]
and BC = 8 cm
BE + EC = 8
y + z = 8
12 - x + z = 8 [From 1]
z = x – 4…. [2]
and
AC = 10 cm
AF + CF = 10
x + z = 10 [From 2]
x + x - 4 = 10
2x = 14
x = 7 cm
Putting value of x in [1] and [2]
y = 12 - 7 = 5 cm
z = 7 - 4 = 3 cm
So, we have AD = 7 cm, BE = 5 cm and CF = 3 cm