The sides AB, BC and AC of a $$\triangle$$ABC are 12 cm, 8 cm and 10 cm respectively. A circle is inscribed in the triangle touching AB, BC and AC at D, E and F, respectively. The difference between the lengths of AD and CE is:

According to the question, we draw the diagram is given below 

AB= 12cm , BC= 8cm ,AC = 10cm

Now taking as an external point

thus AD=AF

$$Let AD =  AF= x cm $$

so, BD= $$ 12-x $$, CE = $$8-(12-x)$$ and CF = $$10-x$$

then 12+8+10= $$x+x+12-x+12-x-4+x-4+x $$

30 = $$2x +16 $$

$$$ 2x= 14 $$

$$x= 7 $$

AD= 7 cm

CE = $$ 8 - (12-7)$$

CE= 8-5 = 3 cm

then difference between  AD and CE is 7-3 = 4 cm

therefore Option (A) 4 cm Ans