Let there is a circle having centre O touches the sides AB and AC of the triangle at point E and F respectively.Let the length of the line segment AE is x.Now inΔABC,CF = CD = 6 (tangents on the circle from point C)BE = BD = 6 (tangents on the circle from point B)AE = AF = x (tangents on the circle from point A)Now AB = AE + EB=> AB = x + 8BC = BD + DC=> BC = 8+6 = 14CA = CF + FA=> CA = 6 + xNows = (AB + BC + CA )/2=> s = (x + 8 + 14 + 6 +x)/2=> s = (2x + 28)/2=> s = x + 14Area of the ΔABC =√{s*(s-a)*(s-b)*(s-c)} =√[(14+x)*{(14+x)-14}*{(14+x)-(6+x)}*{(14+x)-(8+x)}] =√[(14+x)*x*8*6] =√[(14+x)*x*2*4*2*3]=> Area of the ΔABC = 4√[3x(14+x)] .............................1Now area ofΔOBC = (1/2)*OD*BC =(1/2)*4*14 = 56/2 = 28Area ofΔOBC = (1/2)*OF*AC =(1/2)*4*(6+x) = 2(6+x) = 12 + 2xArea ofΔOAB = (1/2)*OE*AB =(1/2)*4*(8+x) = 2(8+x) = 16 + 2xNow Area of the ΔABC = Area ofΔOBC +Area ofΔOBC +Area ofΔOAB=>4√[3x(14+x)] = 28 + 12 + 2x + 16 + 2x=>4√[3x(14+x)] = 56 + 4x=>4√[3x(14+x)] = 4(14 + x)=>√[3x(14+x)] = 14 + xOn squaring bothe side, we get3x(14 + x) = (14 + x)2=> 3x = 14 + x (14 + x = 0 => x = -14 is not possible)=> 3x - x = 14=> 2x = 14=> x = 14/2=> x = 7HenseAB = x + 8=> AB = 7+8=> AB = 15AC = 6 + x=> AC = 6 + 7=> AC = 13So value of AB is 15 cm and value of AC is 13 cm