Step-by-step explanation:
To get to the desired conclusion, you will want to make use of CPCTC and the fact that triangle AXB is congruent to triangle CYD. In order to show those two right triangles are congruent, it is sufficient to claim congruence of their hypotenuses and one internal acute angle (HA congruence). The corresponding internal angles are alternate interior angles where a transversal (BD) crosses parallel lines (AB║CD). It's tedious, but not terribly difficult.
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Statement . . . . Reason
1. Congruent segment pairs are (AB, CD), (AD, BC) . . . . given
2. Perpendicular segment pairs are (AX, BD), (CY, BD) . . . . given
3. ABCD is a parallelogram . . . . definition of parallelogram (opposite sides are the same length)
4. AB ║ CD . . . . definition of parallelogram (opposite sides are parallel)
5. ∠ABX ≅ ∠CDY . . . . alternate interior angles at BD
6. ΔAXB and ΔCYD are right triangles . . . . definition of right triangle (angles at X, Y are 90°)
7. ΔAXB ≅ ΔCYD . . . . HA congruence postulate (AB, CD are the hypotenuses)
8. AX ≅ CY . . . . CPCTC
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Additional comment
A lot of proof has to do with working backward, answering the question, "How do I know ...?" The beginning paragraph in this answer attempts to show you the answers to those questions.
This proof works with the smaller triangles AXB and CYD. You could do the same proof with the larger triangles AXD and CYB. For that, you'd have to claim that AD║BC.