Proof Involving the Diagonals of Parallelogram

No need to do that, but if you feel confident about a solution to a problem just don't post it. Some moderators say that you are not doing a good use of PF if you post easy (for them)...In summary, the conversation revolved around finding the midpoint of diagonals in order to prove that they intersect at their midpoints. The equations of the diagonals were discussed, and it was suggested to use precalculus skills such as understanding equations of a line to solve the problem. It was also mentioned that studying precalculus and calculus 1 at the same time may be challenging, but reducing the number of questions posted per week may help with this.
  • #1
nycmathguy
Homework Statement
Algebra & Geometry
Relevant Equations
Geometric Proof Using Algebra
Chapter 1, Section 1.1.
Look at the picture. Question 57.

Let me see.

To show this prove, I must find the midpoint of the diagonals. The midpoint of (b, c) and (a, 0) must be the same as the midpoint of (0, 0) and
(a + b, c).

You say?
Screenshot_20210613-184014_Drive.jpg
 
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  • #2
I think it is sufficient to see in the picture as you suggest. Analytically you can consider the equation of the diagonals, for example the equation of the diagonal between ##(b,c)## and ##(a,0)## is ##y=\frac{c}{b-a}x-\frac{ca}{b-a}##. The diagonal between ##(a+b,c)## and ##(0,0)## is ##y=\frac{c}{a+b}x##. Now you can try to solve the system of both finding the intersection point ...
Ssnow
 
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  • #3
Ssnow said:
I think it is sufficient to see in the picture as you suggest. Analytically you can consider the equation of the diagonals, for example the equation of the diagonal between ##(b,c)## and ##(a,0)## is ##y=\frac{c}{b-a}x-\frac{ca}{b-a}##. The diagonal between ##(a+b,c)## and ##(0,0)## is ##y=\frac{c}{a+b}x##. Now you can try to solve the system of both finding the intersection point ...
Ssnow
How did you come up with the two equations given in the information in the diagram?
 
  • #4
nycmathguy said:
Homework Statement:: Algebra & Geometry
This is not a homework statement.
An informative homework statement would be "Show or prove that the diagonals of a parallelogram intersect at their midpoints."
nycmathguy said:
Relevant Equations:: Geometric Proof Using Algebra
Not an equation and not relevant. Relevant equations would include those for the midpoint of a line segment and possibly the equation of a line.
nycmathguy said:
Chapter 1, Section 1.1.
Not useful if we don't know which textbook you're using.
nycmathguy said:
Look at the picture. Question 57.

Let me see.
Stream of thought fluff like "let me see" is not useful. It's better to leave this out, especially since you have done this in several of your threads.
nycmathguy said:
To show this prove, I must find the midpoint of the diagonals. The midpoint of (b, c) and (a, 0) must be the same as the midpoint of (0, 0) and
(a + b, c).
Right, and here is where you could have put your calculations.
nycmathguy said:
 
  • #5
nycmathguy said:
How did you come up with the two equations given in the information in the diagram?
He did it using an equation for a line, of which there are at least three. This would have been useful to include in the "Relevant Equations" section of the homework template.

The picture gives the coordinates of the endpoints of the diagonals, so it's just a matter of plugging these values in.

Here are three different equations for a straight line:
  • Slope-intercept form: ##y = mx + b## -- Use it if you know the slope of the line and its y-intercept.
  • Point-slope form: ##y - y_0 = m(x - x_0)## -- Use it if you know the slope of the line and a point on the line. This is the equation form that @Ssnow used.
  • Standard form: ##ax + by + c = 0## -- Every line, including vertical lines, can be written using this equation form.
 
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  • #6
Mark44 said:
He did it using an equation for a line, of which there are at least three. This would have been useful to include in the "Relevant Equations" section of the homework template.

The picture gives the coordinates of the endpoints of the diagonals, so it's just a matter of plugging these values in.

Here are three different equations for a straight line:
  • Slope-intercept form: ##y = mx + b## -- Use it if you know the slope of the line and its y-intercept.
  • Point-slope form: ##y - y_0 = m(x - x_0)## -- Use it if you know the slope of the line and a point on the line. This is the equation form that @Ssnow used.
  • Standard form: ##ax + by + c = 0## -- Every line, including vertical lines, can be written using this equation form.

I know how to use those equations for simple questions not for anything like this thread.
 
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  • #7
nycmathguy said:
I know how to use those equations for simple questions not for anything like this thread.
Since your goal is to study calculus, that's why I suggested a review of precalc material. Any calculus book will assume that you understand and know how to use equations of a line, as well as a lot of other precalc topics.

If you're having trouble with the problem in this thread, that's an even stronger argument for spending time on precalculus stuff.

Despite all of my criticisms, you're doing a lot of things right, and should be commended for your goal of studying calculus. A lot of what I'm telling you is my attempt to help you get your thoughts organized so you can be successful in your goal. I'm also trying to get you to stretch out a bit and risk making mistakes.
 
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  • #8
Mark44 said:
Since your goal is to study calculus, that's why I suggested a review of precalc material. Any calculus book will assume that you understand and know how to use equations of a line, as well as a lot of other precalc topics.

If you're having trouble with the problem in this thread, that's an even stronger argument for spending time on precalculus stuff.

Despite all of my criticisms, you're doing a lot of things right, and should be commended for your goal of studying calculus. A lot of what I'm telling you is my attempt to help you get your thoughts organized so you can be successful in your goal. I'm also trying to get you to stretch out a bit and risk making mistakes.
I started precalculus again back to chapter 1 or have you not noticed my precalculus questions already posted? I am going to tackle precalculus and calculus 1 at the same time but at a slower pace. I am also going to reduce the number of questions posted per week.
 
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  • #9
nycmathguy said:
I am also going to reduce the number of questions posted per week.
No need to do that, but if you feel confident about a solution to a problem just don't post it. Some moderators say that you are not doing a good use of PF if you post easy (for them) problems.
 
  • #10
nycmathguy said:
I know how to use those equations for simple questions not for anything like this thread.
I don't understand why this is difficult for you. You're given the coordinates of all of the points, so as I said before, it's straightforward to plug them into one of the formulas for the equation of a line. Is the problem that the coordinates are given in variables like a and b rather than specified constants like 2 and 5?

nycmathguy said:
I started precalculus again back to chapter 1 or have you not noticed my precalculus questions already posted?
Of course I noticed -- I've responded to most of your threads.
 
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  • #11
Maybe what @Mark44 forgot to tell you in his nice post #5 is that the slope ##m## of a line that connects points (a,b) and (c,d) is ##m=\frac{d-b}{c-a}##
 
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  • #12
Mark44 said:
I don't understand why this is difficult for you. You're given the coordinates of all of the points, so as I said before, it's straightforward to plug them into one of the formulas for the equation of a line. Is the problem that the coordinates are given in variables like a and b rather than specified constants like 2 and 5?

Of course I noticed -- I've responded to most of your threads.

It is not difficult to solve. It is just difficult to set up or better yet, get started.
 
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  • #13
nycmathguy said:
It is not difficult to solve. It is just difficult to set up or better yet, get started.
This doesn't make sense. If one can't figure out how to set up a problem, there's no chance of solving it.
 
  • #14
Mark44 said:
This doesn't make sense. If one can't figure out how to set up a problem, there's no chance of solving it.
I am talking mainly about applications. There are certain word problems (system of equations, for example) that requires setting up equations in two and sometimes three unknowns. My thread about finding two missing numbers is a perfect example.
 

1. What is a parallelogram?

A parallelogram is a quadrilateral with two pairs of parallel sides. This means that the opposite sides of a parallelogram are equal in length and parallel to each other.

2. What is the relationship between the diagonals of a parallelogram?

The diagonals of a parallelogram bisect each other, meaning they intersect at their midpoints. This also means that the diagonals are equal in length.

3. How can you prove that the diagonals of a parallelogram bisect each other?

There are multiple ways to prove this. One way is to use the properties of parallel lines and alternate interior angles. Another way is to use the midpoint formula to show that the coordinates of the intersection point of the diagonals are the same.

4. Can you use the Pythagorean Theorem to prove properties of parallelograms?

Yes, the Pythagorean Theorem can be used to prove that the opposite sides of a parallelogram are equal in length. This is because a parallelogram can be split into two congruent right triangles, and the Pythagorean Theorem can be applied to each triangle to show that the sides are equal.

5. Are there any special cases of parallelograms where the diagonals do not bisect each other?

No, the diagonals of a parallelogram always bisect each other regardless of the size or shape of the parallelogram. This is a property that is unique to parallelograms and does not apply to other quadrilaterals.

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