Answer :

**To Find: AB and AC**

Let the given circle touch the sides AB and AC of the triangle at point E and F respectively and the length of the line segment AF be *x.*

In ∆ABC,

From the theorem which states that the lengths of two tangents drawn from an external point to a circle are equal.So,

CF = CD = 6cm (Tangents on the circle from point C)

BE = BD = 8cm (Tangents on the circle from point B)

AE = AF = *x* (Tangents on the circle from point A)

AB = AE + EB = *x* + 8

BC = BD + DC = 8 + 6 = 14

CA = CF + FA = 6 + *x*

where, a, b and c are sides of triangle and s is semi perimeter.

2*S =* AB + BC + CA

= *x* + 8 + 14 + 6 + *x*

= 28 + 2*x*

*S* =

=√(14 + x)(x)(8)(6)

Also, area of triangle =

And from the given figure it is clear that

Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB

Area of ΔOBC =

Area of ΔOCA =

Area of ΔOAB =

Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB

⇒

⇒

⇒

⇒

⇒ 3x = 14 + x

⇒ 2x = 14

⇒ x = 7

Therefore, *x* = 7

Hence,

AB = *x* + 8 = 7 + 8 = 15 cm

CA = 6 + *x* = 6 + 7 = 13 cm

Hence, AB = 15 cm and AC = 13 cm

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