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

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