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.

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Solution

We know that tangent segments to a circle from the same external point are congruent.

Now, we have

AD = AF, BD = BE and CE = CF

Now,
AD + BD = 14 cm.....(1)

AF + FC = 12 cm

=> AD + FC = 12cm.....(2)

BE + EC = 8 cm

=>BD + FC = 8cm ......(3)

Adding all these we get

AD + BD + AD + FC + BD + FC = 34

=>2(AD + BD + FC) = 34

=>AD+BD+FC=17 cm....(4)

Solving (1) and (4), we get

FC = 3 cm

Solving (2) and (4), we get

BD = 5 cm = BE

Solving (3) and (4), we get:

AD = 9 cm.

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