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Proving the converse is easy enough: you can construct a line between vertices of a parallelogram and then show the resulting triangles are congruent because of adjacent interior angles and ASA congruence of triangles.

However, doesn't the mere existence of a parallelogram imply the parallel postulate? I'm not really sure how to approach this.

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It depends what you mean by a parallelogram. Suppose you have three sides of a parallelogram, defined as a figure with four sides, the opposite pairs being parallel (not meeting when extended indefinitely in either direction). What happens when you try to construct the fourth side?

You need a parallel to the "opposite" side. Pick a point for the fourth side to pass through (eg a presumed vertex). If there is a unique parallel, all is well. But what if there is more than one? You need to show that in this situation the congruence of the parallels means that there is only one, and hence implies the parallel postulate.

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