Parallelogram With non-90 angles

I realized that in-case of a parallelogram with a non-90 degree angles the two diagonals would never be equal. Is my assumption correct ?

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The diagonals of a parallelogram bisect each other, so if the diagonals have equal length, then they cut the parallelogram into four isosceles triangles: a "top and bottom" pair with base angles (say) $\alpha$ and a "left and right" pair with base angles $\beta$. Each corner of the parallelogram has size $\alpha+\beta$, so they're all congruent, which makes them necessarily right angles.

Contrapositively, if the angles aren't right angles, then the diagonals have unequal length.

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So I believe this also means that a diagonal(s) splits a parallelogram (no mater what type they are) into congruent triangles. –  MistyD Jul 3 '12 at 19:15
@MistyD: Correct, in more ways than one. Each diagonal alone cuts a parallelogram into a pair of congruent triangles. (We usually pass through this fact on the way to proving parallelogram properties such as "opposite sides are congruent", "opposite angles are congruent", and, ultimately, that "diagonals bisect each other".) Both diagonals together create four triangles in two congruent pairs. –  Blue Jul 3 '12 at 19:31
Yes, you are correct. Let $ABCD$ be the rhombus. Assume that $AC = BD$. Then the triangles $ABD$ and $ABC$ are congruent (all sides equal). Hence the angle at $A$ equals the angle at $B$. Similarly $\angle C = \angle D$. Since you have a rhombus, $\angle D = \angle B$, so summing up all four angles are equal: $$\angle A = \angle B = \angle C = \angle D$$ and thus they have to be straight.