A triangle ABC is drawn to …

Firstly, consider that the given circle will touch the sides AB and AC of the triangle at point E and F respectively. 

Let AF = x

Now, in ABC,

CF = CD = 6cm  (Tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point C)

BE = BD = 8cm  (Tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point B)

AE = AF = x  (Again, tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point A)

Now, AB = AE + EB 

= x + 8

Also, BC = BD + DC = 8 + 6 = 14 and CA = CF + FA = 6 + x

Now, we get all the sides of a triangle so its area can be find out by using Heron's formula as:

2s = AB + BC + CA

= x + 8 + 14 + 6 + x

= 28 + 2x

⇒ Semi-perimeter = s = (28 + 2x)/2 = 14 + x

Again, area of triangle is also equal to the . Therefore,

Area of ΔOBC = 

Area of ΔOCA =

Area of ΔOAB =

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

On squaring both sides, we get

Either x+14 = 0 or x − 7 =0

Therefore, x = −14and 7

However, x = −14 is not possible as the length of the sides will be negative.

Therefore, x = 7

Hence, AB = x + 8 = 7 + 8 = 15 cm

CA = 6 + x = 6 + 7 = 13 cm