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

\equiv

YM, and point K on AB and N on XY so that BK

\equiv

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

\equiv

NM.

I already know that BK

\equiv

NY and LB

\equiv

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.