Do a parallelograms diagonals bisect each other?

Answer 1

I can't offer a full proof, but I can suggest some possibilities that will lead you to your proof. In a parallelogram, you can easily demonstrate that the angles formed by a cord extending between parallel lines and the parallel lines themselves, and that are formed on opposite sides of the cord, are equal. This will work for both pairs of triangles in the parallelogram, and can be applied to all of the angles at the corners of the parallelogram. This will lead you to demonstrating that the pairs of triangles "pointing" to each other (not adjacent pairs) are similar, and in fact congruent. From there it is not difficult to establish that the connected sections of the two interior cords are equal.

Answer 2

Draw parallelogram ABCD and their diagonals AC and BD intersecting at point E.

Because ABCD is a parallelogram, opposite sides AB and CD are parallel and equal.

Because AB and CD are parallel, angles BDC and ABD are equal. For the same reason, angles ACD and BAC are equal.

Given: AB = CD, angles BDC = ABD and angles ACD and BAC are equal, triangles ABE and CDE are congruent.

Because triangles ABE and CDE are congruent, AE = CE and BE = DE. QED.

Answer 3

The statement cannot be proved because it is not true.

As a counter example, consider a parallelogram which is not a rhombus. Its diagonals do bisect each other but, by definition, it is not a rhombus.

Answer 4

No, they would make an X shape instead of + shape

Answer 5

Draw a parallelogram and cut it out. Then cut from one corner to the other. You should end up with two triangles.

Answer 6

You cannot since, in general, the statement is not true.

Answer 7

Yes, they do.

Answer 8

yes

Answer 9

yes they do