### 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 the sine of their respective angles:

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

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

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

Simplifying, we find:

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

Given that $BD\sin \angle ABC = a\sin C$, it follows directly that:

$$BD = b$$

Thus, we have proven that $BD = b$.

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

#### Method 1:

From part (1), we know $BD = b$. Given $AD = 2DC$, 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 $\triangle CBD$:

- In $\triangle ABD$:

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

- In $\triangle CBD$:

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

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

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

Solving this equation, we find:

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

Given $b^{2} = ac$, we substitute and solve for $c$:

$$3c^{2} - 11ac + 6a^{2} = 0 \implies c = 3a \text{ or } c = \frac{2}{3}a$$

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

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

- For $c = 3a$, we discard $\cos \angle ABC = \frac{7}{6}$ as it's not possible.
- For $c = \frac{2}{3}a$, we find:

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

Therefore, we conclude that $\boxed{\cos \angle ABC = \frac{7}{12}}$.