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
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 $
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
So, angles KBL and NYM are equal and so are angles ABC and