# 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.
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.