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Question
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= 12cm, BC=8cm and AC = 10cm, find the length of AD, BE and CF.
Solution
Sol:
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 = l2cm …….(1)
AF + FC = l0 cm
⇒ AD + FC = l0 cm …….(2)
BE + EC = 8 cm
⇒ BD + FC = 8cm …….(3)
Adding all these we get
AD + BD + AD + FC + BD + FC = 30
⇒ 2(AD + BD + FC) = 30
⇒ AD + BD + FC = l5cm …….(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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