1. Identify the given conditions and setup:
- In the convex quadrilateral $ABCD$, $\angle ABC = \angle ADC = 90^\circ$.
- Point $P$ is chosen on diagonal $BD$ such that $\angle APB = 2\angle CPD$.
- Points $X$ and $Y$ are chosen on segment $AP$ such that $\angle AXB = 2\angle ADB$ and $\angle AYD = 2\angle ABD$.

2. **Introduce the circumcenter $O$ and point $Q$:**
- Let $O$ be the circumcenter of $ABCD$.
- Let $Q$ be the second intersection of line $AP$ with the circumcircle $(ABCD)$.

3. Identify the intersections:
- Points $X$ and $Y$ are the second intersections of $AP$ with the circumcircles $(AOD)$ and $(AOB)$, respectively.

4. Establish the similarities:
- $\triangle DOY \sim \triangle DCQ$:
- $\angle DYO = \angle DAO = \angle DAC = \angle DQC$
- $\angle YDO = \angle YAO = \angle QAC = \angle QDC$
- $\triangle OYX \sim \triangle CBD$:
- $\angle OYX = \angle ODA = \angle OAD = \angle CAD = \angle CBD$
- $\angle OXY = \angle OBA = \angle OAB = \angle CAB = \angle CDB$

5. Use the similarities to derive ratios:
- From $\triangle DOY \sim \triangle DCQ$:
$$\frac{R}{OY} = \frac{CD}{CQ}$$
- From $\triangle OYX \sim \triangle CBD$:
$$\frac{OY}{XY} = \frac{BC}{BD}$$

6. Combine the ratios:
$$\frac{OY}{XY} \cdot \frac{R}{OY} = \frac{CD}{CQ} \cdot \frac{BC}{BD}$$

7. **Compute $CQ$ using the Law of Sines in $\triangle CDP$:**
- Using the Law of Sines:
$$\frac{CP}{\sin \angle BDC} = \frac{CD}{\sin \angle CPD} = \frac{CD}{\sin \angle CPQ}$$
- Therefore, $CQ = CD \cdot \sin \angle BDC$.

8. **Substitute $CQ$ and simplify:**
$$\frac{1}{XY} = \frac{1}{R} \cdot \frac{CD}{CQ} \cdot \frac{BC}{BD} = \frac{1}{R \cdot \sin \angle BDC \cdot \frac{BD}{BC}} = \frac{1}{\frac{1}{2} \cdot BD}$$

9. Conclude the proof:
$$XY = \frac{BD}{2}$$

The final answer is $\boxed{ BD = 2XY }$.