Write an indirect proof to show that opposite sides of a parallelogram are congruent. Be sure to create and name the appropriate geometric figures. This figure does not need to be submitted.

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Write an indirect proof to show that opposite sides of a parallelogram are congruent. Be sure to create and name the appropriate geometric figures. This figure does not need to be submitted.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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@johnweldon1993 can u please help me !!!
@mathmate please please help me
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no all it telling is to Write an indirect proof to show that opposite sides of a parallelogram are congruent
Ok. To start, congruent opposite sides of a parallelogram is one of the properties of a parallelogram. Since we need to prove it, we have to start with the definition of a parallelogram to know what we already know. According to Wiki: "In Euclidean geometry, a parallelogram is a (non self-intersecting) quadrilateral with two pairs of parallel sides. " Are you familiar with congruent triangles, and transversals of parallel lines?
yes I am (:
Good, so this is what we're given with. |dw:1433891186196:dw| We need to prove that AD // BC
* we need to prove that mAD = mBC
@Moo_Moo17 What do you think the approach would be?
im not sure I feel like I know it but I cant remember sorry im trying
|dw:1433891607737:dw| The trick is to draw the diagonal BD, then you could prove that triangles ABD and CDB are congruent. Since AD and BC are corresponding sides, they would be congruent. I'll let you attempt that, since you are familiar with the tools needed (parallel lines, transversals, congruent triangles).
@mathmate im some what fimilar with the tools really sorry
|dw:1433892491187:dw| You would prove congruence by ASA. Look up your notes to find the justifications for angle ABC = angle CDB ....[justification] BD=BD ....[justification] angle BDA = angle DBC ....[justification] Therefore triangles ABD and CDB are congruent [ASA, or angle-side-angle] and mAB=mCD [corresponding sides of congruent triangles]. You can use the following link as a review. http://www.mathsisfun.com/geometry/triangles-congruent-finding.html
oh ok the answer would triangles ABD and CDB are congruent [ASA, or angle-side-angle] and mAB=mCD [corresponding sides of congruent triangles].
You still need 3 justifications that I left blank to complete the proof.
oh ok so so the 1 is ASA I think 2 m not to sure what it would be 3 SSS I think
You only have to fill in the red parts to complete the proof. The rest is already done for you. Use the link if necessary to find the justifications: http://www.mathsisfun.com/geometry/triangles-congruent-finding.html angle ABC = angle CDB ....[\(\color{red}{justification}\)] BD=BD ....[\(\color{red}{justification}\)] angle BDA = angle DBC ....[\(\color{red}{justification}\)] Therefore triangles ABD and CDB are congruent [ASA, or angle-side-angle] and mAB=mCD [corresponding sides of congruent triangles].
im trying to figure it out but im having trouble can u please help me I don't understand
|dw:1433895706289:dw| When a transversal cuts two parallel lines, it makes many congruent angles, out of which: Vertically opposite angles are: a=d, b=c, f=g, e=h corresponding angles a=e, c=g b=f, d=h alternate interior angles c=f, d=e Alternate exterior angles a=h, b=g It seems a lot, but if you study it, they are all logical groups. If you work with those, you will find the justifications not too hard to find.
ok so ABC = CDB are Vertically opposite angles angle BDA = DBC is Corresponding angles Angle BD= BD is think alternate interior angles
Sorry, not quite. You need to see how the groups work. BD is not even an angle. BTW, have you done geometric proofs before? It seems that you're quite rusty about it.
I sorta have I wasn't very good at it
It would be a good idea to study and understand the link I gave you, and then you'll be better at it. No efforts, no results. http://www.mathsisfun.com/geometry/triangles-congruent-finding.html
and this one for congruent angles: https://www.mathsisfun.com/geometry/parallel-lines.html
idk i just idk im sorry im so confused im very good geometry iv studied and studied but im not good at anything
|dw:1433900762255:dw| Consider the above diagram.
@Moo_Moo17 you there?
yea i am sorry
No sorry, it took me a while to come back, sorry about that.
Seeing that BC is parallel to AD, what can you say about angles ADB and DBC?
They are equal, because they are alternate interior angles. (read the second link)
Here is a guide to show you that one pair of sides are congruent. I leave the rest of the proof for you to do on your own. You'll have similar steps to mine that I show in the attached pdf.
BC is congruent to AD but also i think they are corresponding angles or transversal
AD is equal to AD with the justification "common", because it's the same length.

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