Question

In Fig. 2, a circle is inscribed in a ΔABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. If the lengths of sides AB, BC and CA and 12 cm, 8 cm and 10 cm respectively, find the lengths of AD, BE and CF.

Answer 1

t is given that

AB = 12 cm

⇒ AD + BD = 12 cm    .....(1)

BC = 8 cm

⇒ BE + CE = 8 cm       .....(2)

CA = 10 cm

⇒ AF + CF = 10 cm     .....(3)

CF and CE act as tangents to the circle from the external point C.

It is known that the lengths of tangents drawn from an external point to a circle are equal.

∴ CF = CE                   .....(4)

Similarly, AF and AD act as tangents to the circle from the external point A

∴ AF = AD                  .....(5)

Also, BD and BE act as tangents to the circle from the external point B.

∴ BD = BE                  .....(6)

Using (4) and (2), we get

BE + CF = 8 cm          .....(7)

Using (5) and (3), we get

AD + CF = 10 cm       .....(8)

Using (6) and (1), we get

AD + BE = 12 cm       .....(9)

Adding (7), (8) and (9), we get

BE + CF + AD + CF + AD + BE = 8 cm + 10 cm + 12 cm

⇒ 2AD + 2BE + 2CF = 30 cm

⇒ 2(AD + BE + CF) = 30 cm

⇒ AD + BE + CF = 15 cm .....(10)

Subtracting (7) from (10), we get

AD + BE + CF − BE − CF = 15 cm − 8 cm

⇒ AD = 7 cm

Subtracting (8) from (10), we get

AD + BE + CF − AD − CF = 15 cm − 10 cm

⇒ BE = 5 cm

Subtracting (9) from (10), we get

AD + BE + CF − AD − BE = 15 cm − 12 cm

⇒ CF = 3 cm

Thus, the lengths of AD, BE and CF are 7 cm, 5 cm and 3 cm, respectively.