Problem 1498

1. Angle Chasing:
Given the problem, we start by defining the angles:
- Let $\angle DBC = \alpha$.
- Then $\angle DAB = 2\alpha$.
- Given $\angle ADB - \angle ABD = 4\alpha$, we can write:
$$\angle ADB = 90^\circ + \alpha \quad \text{and} \quad \angle ABD = 90^\circ - 3\alpha.$$
- Since $\angle ADC = 135^\circ$, we have:
$$\angle BDC = 45^\circ - \alpha.$$
- Also, $\angle DCB = 135^\circ$.

2. Intersection Point:
Let $BC \cap AD = \{F\}$. Note that $\triangle FDC$ is right-angled at $F$ with one angle of $45^\circ$. This implies:
$$\angle DFC = 45^\circ.$$

3. Trigonometric Relationships:
We need to evaluate $\sin \alpha$ and $\cos \alpha$. Given $BC = x$ and $BC = \sqrt{2} CD$, let $CD = y$. Then:
$$x = \sqrt{2} y.$$

4. Using Right Triangle Properties:
In $\triangle FDC$, since $\angle DFC = 45^\circ$, we have:
$$\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}.$$
Therefore, $FD = FC = \frac{y}{\sqrt{2}}$.

5. **Finding $AB$ and $AD$:**
- To find $AB$ and $AD$ in terms of $x$:
$$AD = AF + FD = AF + \frac{y}{\sqrt{2}}.$$
- Since $BC = x$ and $BC = \sqrt{2} y$, we have:
$$y = \frac{x}{\sqrt{2}}.$$
- Thus:
$$FD = \frac{y}{\sqrt{2}} = \frac{\frac{x}{\sqrt{2}}}{\sqrt{2}} = \frac{x}{2}.$$

6. Using the Given Condition:
- We need to show $AB = BC + AD$.
- From the problem, we know $BC = x$ and $AD = AF + \frac{x}{2}$.
- To find $AB$, we use the fact that $\angle DAB = 2\alpha$ and $\angle ABD = 90^\circ - 3\alpha$.

7. Conclusion:
- By the given conditions and the trigonometric relationships, we can conclude that:
$$AB = BC + AD.$$
- Therefore, the given condition holds true.

The final answer is $AB = BC + AD$.