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I need to check my answer!
Let's assume that the diagonals of a parallelogram do not bisect one another. A parallelogram has four sides and it's opposite angles and sides are congruent and parallel. If it's opposite sides are parallel and congruent, and so are it's angles, then the diagonals of a parallelogram must intersect each other.
Seems that here you are proving that they intersect and not that they bisect
What's the difference between intersection and bisection?
I thought they were the same. :/
so intersection is that two lines meet at the same point bisection is that they meet at the same point AND this point splits them in two equal segments
Wait, I think I remember, bisection means they cut eachothe rin halves
Ahh, okay! I forgot completely.
But the start of your proof seemed reasonable, you should elaborate further
Remind: Should state that parallelogram is not a reflex quadrilateral, hence their diagonals intersect.
Thank you, I'm understanding better now. But do you think you could give me a hint to get started on when else I should be writing, please?
Is this one okay? Let's assume that the diagonals of a parallelogram do not bisect one another. A parallelogram has four sides and it's opposite angles and opposite sides are congruent and parallel. If the diagonals of a parallelogram intersect each other, and a parallelogram's opposite angles and opposite sides are congruent and parallel, then the intersecting diagonals must also be bisecting one another.