Nice. You can alternatively prove the negative: suppose it's a parallelogram with non-right angles, assume angle at A is "sharper". Thus, you can create two right angle triangles: AB'C with the base extended beyond B (= B') where AC is one diagonal and A'BD where A' is "below" D and BD is the other diagonal. As AB'C is larger than A'BD (because A' is within AB and B' is outside while the height is the same) AC can't be the same length as BD.

Thus, no parallelogram with non-right angles exists that has equal length diagonals.