A circle touches the four sides of a quadrilateral ABCD. The value of $$\frac{AB+CD}{CB+DA}$$ is equal to:

Circle touches the four sides of quadrilateral ABCD as shown in the figure

AT and AP are tangents to the circle from point A

$$=$$>  AT = AP

Let AT = AP = w

BP and BS are tangents to the circle from point B

$$=$$>  BP = BS

Let BP = BS = x

CS and CR are tangents to the circle from point C

$$=$$>  CS = CR

Let CS = CR = z

DR and DT are tangents to the circle from point D

$$=$$>  DR = DT

Let DR = DT = y

$$\therefore\ $$ $$\frac{AB+CD}{CB+DA}$$ = $$\frac{w+x+z+y}{z+x+y+w}$$ = 1

Hence, the correct answer is Option A