Proving a parallelogram is a rhombus, given an Isosceles triangle

In summary: I'm having a problem with the fact that we find this congruence:ST ≅ RUAlthough the instructor forced that congruence into the statement located directly below. The only reason I have those boxes prefilled is because they were done so by the instructor. Maybe this doesn't break what your saying, and I don't think it does. What you state is perfectly logical from my understanding, however, the instructor seems to have found some transitive property just before the "ST ≅ RU" statement. If you understand why this is done, could you please explain it to me. Is he simply utilizing a reflexive property to state what is already stated through the transitive property above
  • #1
Joodez
12
0
I believe I could work this problem fine had the instructor not placed statements #3,4, and 6, as well as the reasons # 5 and 7. I can't seem to understand where he's using the transitive property in #5. If these steps weren't here I would assume, I'm only using substitution to prove that "RS" = "RU" and therefore satisfies, along with the parallelogram statement, that this quadrilateral or parallelogram is indeed a rhombus. Coming from the statement of a parallelogram and knowing the congruency of two of the adjacent sides, I just don't understand why there's any need for a transitive property. Does anyone have any ideas or a direction they can lead me in?

20160214_161326.jpg
 
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  • #2
Joodez said:
I believe I could work this problem fine had the instructor not placed statements #3,4, and 6, as well as the reasons # 5 and 7. I can't seem to understand where he's using the transitive property in #5. If these steps weren't here I would assume, I'm only using substitution to prove that "RS" = "RU" and therefore satisfies, along with the parallelogram statement, that this quadrilateral or parallelogram is indeed a rhombus. Coming from the statement of a parallelogram and knowing the congruency of two of the adjacent sides, I just don't understand why there's any need for a transitive property. Does anyone have any ideas or a direction they can lead me in?

20160214_161326.jpg
Your image doesn't show in the thread.

In your own words, how do you state the transitive property?
 
  • #3
The transitive property of equality states that: a=b, b=c, therefore; a=c

3vfoKCn.jpg


direct link: http://i.imgur.com/3vfoKCn.jpg

I uploaded the image to a different provider. Let me know if there are still any issues. I would write down the statements and reasons, but without the rhombus picture, it may be hard to visualize what's going on.
 
  • #4
Joodez said:
The transitive property of equality states that: a=b, b=c, therefore; a=c

3vfoKCn.jpg


direct link: http://i.imgur.com/3vfoKCn.jpg

I uploaded the image to a different provider. Let me know if there are still any issues. I would write down the statements and reasons, but without the rhombus picture, it may be hard to visualize what's going on.
I could see it OK in my reply panel, but not directly in your post.

The transitive property should hold equally well for congruence, so what's wrong with using it as follows:

ST ≅ RU , RU ≅ UT , ∴ ST ≅ UT
 
  • #5
Given x=y and y=z, what can you deduce? Does that not depend on the transitivity of equality!
By the way, I'm not aware that triangle RST being isosceles implies a specific two sides are equal, so I don't know how we get statement 3. Could it not be RS=RT?
 
  • #6
SammyS said:
I could see it OK in my reply panel, but not directly in your post.

The transitive property should hold equally well for congruence, so what's wrong with using it as follows:

ST ≅ RU , RU ≅ UT , ∴ ST ≅ UT
I'm having a problem with the fact that we find this congruence:

ST ≅ RU

Although the instructor forced that congruence into the statement located directly below. The only reason I have those boxes prefilled is because they were done so by the instructor. Maybe this doesn't break what your saying, and I don't think it does. What you state is perfectly logical from my understanding, however, the instructor seems to have found some transitive property just before the "ST ≅ RU" statement. If you understand why this is done, could you please explain it to me. Is he simply utilizing a reflexive property to state what is already stated through the transitive property above?
 
  • #7
haruspex said:
Given x=y and y=z, what can you deduce? Does that not depend on the transitivity of equality!
By the way, I'm not aware that triangle RST being isosceles implies a specific two sides are equal, so I don't know how we get statement 3. Could it not be RS=RT?
Yipes, thanks for pointing this out. Only two sides of an isosceles triangle "must" be equivalent. Alright, I'm going to have to rework this from there on a bit and see what I come out with. Sorry for posting incorrect calculations in the problem above, in fact, let me post the original problem from the worksheet so as not to derail anyone.

1MzZelo.png
 
  • #8
Joodez said:
Yipes, thanks for pointing this out. Only two sides of an isosceles triangle "must" be equivalent. Alright, I'm going to have to rework this from there on a bit and see what I come out with. Sorry for posting incorrect calculations in the problem above, in fact, let me post the original problem from the worksheet so as not to derail anyone.

1MzZelo.png
Seems to me that you have to assume the case RS=ST or it is not necessarily true that the figure is a rhombus. So go with your original attempt and just fill in the remaining blanks. It's just a poorly specified question, or maybe there's some convention I don't know whereby "ABC is isosceles" implies that AB and BC are equal.
 
  • #9
Alright, so now that I realize that this:
RS ≅ RT
is just as possible with the given information as this:
RS ≅ ST
and this:
ST ≅ RT
so...
how can it be proven at all that this object is a rhombus? Because it is a parallelogram, side RS ≅ UT and ST ≅ RU. This is given, based upon the formula for a parallelogram instead of that of the Isosceles.

From here, I don't realize how a transitive property statement is eligible to be made. If "ST ≅ RT", then how can we be certain that side ST will ever be congruent to side RS?
 
  • #10
haruspex said:
Seems to me that you have to assume the case RS=ST or it is not necessarily true that the figure is a rhombus. So go with your original attempt and just fill in the remaining blanks. It's just a poorly specified question, or maybe there's some convention I don't know whereby "ABC is isosceles" implies that AB and BC are equal.
This is exactly the issue I see with the way this problem is formatted. I too am unaware as to the reasoning behind "triangle" RST defining RS ≅ ST. This just doesn't add up to me.

If I am to assume the original "incorrect" train of thought, what statement could the teacher be making in #5 to provide the transitive property reason? I think he's inferring this:

ST ≅ UT, because RS ≅ ST and RS ≅ UT

I see the transitive property being used for that statement, I guess, but after the fact that I can't solidify which sides are which for that isosceles, I don't really follow the current train of thought. To me, it seems like the entirety of this statement is broken unless, the format "triangle" RST means S is the vertex at which RS and ST are declared ≅.
 
  • #11
Joodez said:
Alright, so now that I realize that this:
RS ≅ RT
is just as possible with the given information as this:
RS ≅ ST
and this:
ST ≅ RT
so...
how can it be proven at all that this object is a rhombus? Because it is a parallelogram, side RS ≅ UT and ST ≅ RU. This is given, based upon the formula for a parallelogram instead of that of the Isosceles.

From here, I don't realize how a transitive property statement is eligible to be made. If "ST ≅ RT", then how can we be certain that side ST will ever be congruent to side RS?
upload_2016-2-14_18-48-15.png


The instructor's reason given for #3 is that ΔRST is isosceles. Yes, it should have been stated in the given which two sides are congruent. However, wasn't that indicated in the figure?
 
  • #12
SammyS said:
View attachment 95844

The instructor's reason given for #3 is that ΔRST is isosceles. Yes, it should have been stated in the given which two sides are congruent. However, wasn't that indicated in the figure?
It is not indicated in the printed version in post #7.
 
  • #13
There's at least a hint of it here from the original figure.
upload_2016-2-14_19-30-59.png
 
  • #14
SammyS said:
There's at least a hint of it here from the original figure.
View attachment 95853
Sure, but I presumed that was from Joodez' interpretation of the information.
 
  • #15
haruspex said:
Sure, but I presumed that was from Joodez' interpretation of the information.
Aye, that was my original (now taken back) interpretation of what I was actually considering to be an "equilateral" triangle.

Below is what I found to be the finished proof, assuming that the isosceles triangle given means that RS = ST.

KNAUv8k.jpg
 
  • #16
My browser needed updating & was giving me fits, so I missed some of these posts. I see now.
 
  • #17
SammyS said:
My browser needed updating & was giving me fits, so I missed some of these posts. I see now.
No worries, it happens to the best of us. Thank you to you, SammyS, and haruspex. You helped me out a lot here, by pointing out the things you did.
 

Related to Proving a parallelogram is a rhombus, given an Isosceles triangle

What is a parallelogram?

A parallelogram is a quadrilateral with two pairs of parallel sides.

What is a rhombus?

A rhombus is a parallelogram with four equal sides.

How can an isosceles triangle prove a parallelogram is a rhombus?

An isosceles triangle has two equal sides and two equal angles. If the equal sides are opposite angles of a parallelogram, then the parallelogram must be a rhombus.

What mathematical properties must be shown to prove a parallelogram is a rhombus?

To prove a parallelogram is a rhombus, we must show that it has two pairs of parallel sides and all four sides are equal in length. Additionally, we must show that the opposite angles are equal.

Can a parallelogram be a rhombus if it does not have two pairs of parallel sides?

No, a parallelogram must have two pairs of parallel sides to be a rhombus. If it does not have two pairs of parallel sides, it is either a rectangle, square, or a general parallelogram.

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