In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at point D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the lengths of AD, BE and CF [CBSE 2013]

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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 = 12 cm .....(1)
AF + FC = 10 cm
⇒ AD + FC = 10 cm .....(2)
BE + EC = 8 cm
⇒ BD + FC = 8 cm .....(3)
Adding all these we get
AD + BD + AD + FC + BD + FC = 30
⇒2(AD + BD + FC) = 30
⇒AD + BD + FC = 15 cm .....(4)
Solving (1) and (4), we get
FC = 3 cm
Solving (2) and (4), we get
BD = 5 cm
Solving (3) and (4), we get
and AD = 7 cm
∴ AD = AF = 7 cm, BD = BE = 5 cm and CE = CF = 3 cm

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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 fo AD, BE and CF.

In Fig., a circle inscribed in triangle ABC touches its sides AB,BC and AC at points D,E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, then find the lengths of AD, BE and CF.