1. Let the common magnitude of the vectors be |vec a|=|vec b|=m.

2. Let the angle between $\vec a$ and $\vec b$ be $\theta$. Then $|\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b=2m^2+2m^2\cos\theta=2m^2(1+\cos\theta).$ Similarly, $|\vec a-\vec b|^2=2m^2(1-\cos\theta).$

3. Using half-angle identities, $1+\cos\theta=2\cos^2\frac\theta2,\qquad 1-\cos\theta=2\sin^2\frac\theta2.$ Hence

   \qquad |\vec a-\vec b|=2m\sin\frac\theta2,$$ taking positive values since these are magnitudes.

4. Given $\frac{|\vec a+\vec b|+|\vec a-\vec b|}{|\vec a+\vec b|-|\vec a-\vec b|}=\sqrt2+1.$ Substitute the above expressions: $\frac{2m\cos\frac\theta2+2m\sin\frac\theta2}{2m\cos\frac\theta2-2m\sin\frac\theta2}=\sqrt2+1.$ Cancelling $2m$, $\frac{\cos\frac\theta2+\sin\frac\theta2}{\cos\frac\theta2-\sin\frac\theta2}=\sqrt2+1.$

5. Put $x=\tan\frac\theta2.$ Dividing numerator and denominator by $\cos\frac\theta2$, $\frac{1+x}{1-x}=\sqrt2+1.$ Solve for $x$: $1+x=(\sqrt2+1)(1-x),$ $1+x=\sqrt2+1-(\sqrt2+1)x,$ $x+(\sqrt2+1)x=\sqrt2,$ $x(\sqrt2+2)=\sqrt2,$ $x=\frac{\sqrt2}{\sqrt2+2}=\sqrt2-1.$ Therefore, $\tan\frac\theta2=\sqrt2-1.$

6. We need $\frac{|\vec a+\vec b|^2}{|\vec a|^2}.$ Since $|\vec a+\vec b|=2m\cos\frac\theta2,$ we get \frac{|\vec a+\vec b|^2}{|\vec a|^2}=rac{4m^2\cos^2\frac\theta2}{m^2}=4\cos^2\frac\theta2.

7. Now use $\cos^2\frac\theta2=\frac{1}{1+\tan^2\frac\theta2}.$ With $\tan\frac\theta2=\sqrt2-1$, $\tan^2\frac\theta2=(\sqrt2-1)^2=3-2\sqrt2.$ So $\cos^2\frac\theta2=\frac{1}{1+3-2\sqrt2}=\frac{1}{4-2\sqrt2}.$ Hence $4\cos^2\frac\theta2=\frac{4}{4-2\sqrt2}.$ Simplify:

   =\frac{2(2+\sqrt2)}{4-2} =2+\sqrt2.$$

8. Therefore, $\frac{|\vec a+\vec b|^2}{|\vec a|^2}=2+\sqrt2.$

So the correct option is A.