Many thanks Andrei (School 205, Bucharest) for this well explained solution. A correct solution was also sent by Thomas, whilst Lawrence, of Beecroft Primary, may not have managed the complete journey - he certainly saw the importance of the parallel lines!

Let L be a point on BC and M on YZ so that:
BL ≡ YM, and point K on AB and N on XY so that BK ≡ YN.
As BL is parallel and congruent with YM, BLMY is a parallelogram, and LM is parallel and congruent with BY.

In the same situation, BK is parallel and congruent with YN and BKNY is a parallelogram.

So, KN is parallel and congruent with BY.

From the two relations, I observe that KN is parallel and congruent with LM and KLMN is a parallelogram.

So, KL ≡ NM.
I already know that BK ≡ NY and LB ≡ MY, and from these three congruence relations, triangles KBL and NYM are congruent.

So, angles KBL and NYM are equal and so are angles ABC and XYZ.