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Another Grade 7 Math Problem (1 Viewer)

chet

chet

Footballguy
My daughter was asked to prove that the diagonals of a parallelogram bisect each other. This is easy other than she hasn't learned about triangle congruency. Any way to prove that without knowing about triangle congruency?

 
The easiest way is to measure the opposing angles of the parallelogram with a protractor. I seem to remember there being certain rules to proving the existence of a parallelogram, but I can't recall them off of the top of my head.

 
If you cut a parallellogram in half they make two equal triangles. You can prove this because their top angles are identical and the bottom line is identical (the cut you made is just as long on one triangle as the other because it goes from the same point to the same point) and their sides are identical (they already were because a parallellogram has equal length on its parallel sides). Two equal triangles have equal angles. Cutting an angle into two equal angles is called bisecting. QED parellelograms are bisectual.

 
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bostonfred said:
If you cut a parallellogram in half they make two equal triangles. You can prove this because their top angles are identical and the bottom line is identical (the cut you made is just as long on one triangle as the other because it goes from the same point to the same point) and their sides are identical (they already were because a parallellogram has equal length on its parallel sides). Two equal triangles have equal angles. Cutting an angle into two equal angles is called bisecting. QED parellelograms are bisectual.
Click to expand...
I knew that sooner or later, we'd get to threesomes.

 
draw the picture and start proving which inner triangles are equivalent...From ther, it should fall out that the lines bisect each other

 
Rove! said:
draw the picture and start proving discovering which inner triangles are equivalent congruent...From there, it should fall out that the lines bisect each other
Click to expand...
FYP

Aside from the misuse of the terms proving and equivalent, this is how I would attack the problem. That is, assuming we're not looking for a formal geometric proof.

 
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