### Part (1) Proof:

Given that $b^{2}=ac$, we start by applying the Law of Sines in $\triangle ABC$:

$$\frac{b}{\sin \angle ABC} = \frac{c}{\sin \angle ACB} = 2R$$

This gives us expressions for $b$ and $c$ in terms of $R$ and their respective angles:

$$b = 2R\sin \angle ABC, \quad c = 2R\sin \angle ACB$$

Substituting these into the given condition $b^{2}=ac$, we get:

$$(2R\sin \angle ABC)^{2} = a(2R\sin \angle ACB)$$

Simplifying this equation, we find:

$$b\sin \angle ABC = a\sin C$$

Given that $BD\sin \angle ABC = a\sin C$, we can conclude that:

$$BD = b$$

Therefore, we have proved that $BD = b$.

### Part (2) Finding $\cos \angle ABC$:

#### Method 1:

Given $BD = b$ and $AD = 2DC$, we can express $AD$ and $DC$ as fractions of $b$:

$$AD = \frac{2}{3}b, \quad DC = \frac{1}{3}b$$

Applying the Law of Cosines in $\triangle ABD$ and $\triangle CBD$, we get:

$$\cos \angle BDA = \frac{b^{2} + \left(\frac{2}{3}b\right)^{2} - c^{2}}{2b \cdot \frac{2}{3}b} = \frac{13b^{2} - 9c^{2}}{12b^{2}}$$

$$\cos \angle BDC = \frac{b^{2} + \left(\frac{1}{3}b\right)^{2} - a^{2}}{2b \cdot \frac{1}{3}b} = \frac{10b^{2} - 9a^{2}}{6b^{2}}$$

Since $\angle BDA + \angle BDC = \pi$, we have $\cos \angle BDA + \cos \angle BDC = 0$, leading to:

$$\frac{13b^{2} - 9c^{2}}{12b^{2}} + \frac{10b^{2} - 9a^{2}}{6b^{2}} = 0$$

Solving this equation gives us:

$$11b^{2} = 3c^{2} + 6a^{2}$$

Given $b^{2} = ac$, we find:

$$3c^{2} - 11ac + 6a^{2} = 0$$

This yields two possible values for $c$:

$$c = 3a \quad \text{or} \quad c = \frac{2}{3}a$$

Using the Law of Cosines in $\triangle ABC$:

$$\cos \angle ABC = \frac{a^{2} + c^{2} - ac}{2ac}$$

Substituting the values of $c$:

- When $c = 3a$, $\cos \angle ABC = \frac{7}{6}$ (rejected since $\cos$ values must be $\leq 1$).
- When $c = \frac{2}{3}a$, $\cos \angle ABC = \frac{7}{12}$.

Therefore, we conclude that:

$$\boxed{\cos \angle ABC = \frac{7}{12}}$$