Let $$\vec{a}$$ and $$\vec{b}$$ be the vectors of the same magnitude such that $$\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|} = \sqrt{2}+1$$. Then $$\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2}$$ is :

Let the common magnitude of the two vectors be $$|\vec a| = |\vec b| = m$$ and let the angle between them be $$\theta$$.

Using the cosine rule for vectors:
$$|\vec a + \vec b| = \sqrt{m^2 + m^2 + 2m^2 \cos\theta} = m\sqrt{2(1+\cos\theta)}$$
$$|\vec a - \vec b| = \sqrt{m^2 + m^2 - 2m^2 \cos\theta} = m\sqrt{2(1-\cos\theta)}$$

Rewrite these with the half-angle substitution $$\theta = 2x$$ (so $$x = \theta/2$$):
$$|\vec a + \vec b| = 2m\cos x$$
$$|\vec a - \vec b| = 2m\sin x$$

The given ratio becomes
$$\frac{|\vec a+\vec b| + |\vec a-\vec b|}{|\vec a+\vec b| - |\vec a-\vec b|} = \frac{2m(\cos x + \sin x)}{2m(\cos x - \sin x)} = \frac{\cos x + \sin x}{\cos x - \sin x}$$

This ratio equals $$\sqrt{2}+1$$, hence
$$\frac{\cos x + \sin x}{\cos x - \sin x} = \sqrt{2}+1 \; -(1)$$

Cross-multiplying in $$(1)$$:
$$\cos x + \sin x = (\sqrt{2}+1)(\cos x - \sin x)$$
$$\cos x + \sin x - (\sqrt{2}+1)\cos x + (\sqrt{2}+1)\sin x = 0$$
$$\cos x(1-\sqrt{2}-1) + \sin x(1+\sqrt{2}+1) = 0$$
$$\cos x(-\sqrt{2}) + \sin x(\sqrt{2}+2) = 0$$
$$\tan x = \frac{\sqrt{2}}{\sqrt{2}+2} \; -(2)$$

The quantity required is
$$\frac{|\vec a+\vec b|^{2}}{|\vec a|^{2}} = \frac{(2m\cos x)^{2}}{m^{2}} = 4\cos^{2}x \; -(3)$$

From $$(2)$$, set $$\tan x = t = \dfrac{\sqrt{2}}{\sqrt{2}+2}$$. Then
$$\cos^{2}x = \frac{1}{1+t^{2}}$$

Compute $$t^{2}$$:
$$t^{2} = \frac{2}{(\sqrt{2}+2)^{2}} = \frac{2}{6+4\sqrt{2}} = \frac{1}{3+2\sqrt{2}}$$

Hence
$$1+t^{2} = 1 + \frac{1}{3+2\sqrt{2}} = \frac{4+2\sqrt{2}}{3+2\sqrt{2}}$$

Insert this into $$(3)$$:
$$4\cos^{2}x = \frac{4}{1+t^{2}} = 4 \cdot \frac{3+2\sqrt{2}}{4+2\sqrt{2}}$$
$$= \frac{12 + 8\sqrt{2}}{4 + 2\sqrt{2}} = \frac{6 + 4\sqrt{2}}{2 + \sqrt{2}} = 2 + \sqrt{2}$$

Therefore,
$$\frac{|\vec a + \vec b|^{2}}{|\vec a|^{2}} = 2 + \sqrt{2}$$

The correct choice is Option C.