Vector prove that diagonals of rhomb split in ratio 1/2.

In summary, the proof involves using the fact that the diagonals of a rhombus are perpendicular bisectors of each other and setting up equations using vector notation to prove that the diagonals split in half in a 1:2 ratio. It also requires an understanding of linear independence to account for cases where the diagonals may be parallel.
  • #1
borovecm
14
0
Vector proof that diagonals of rhomb split in ratio 1/2.

Homework Statement



Hi. For my math homework I have to prove with vectors(we are currently learning that) that diagonals of any rhomb split in half in ratio 1/2.

Homework Equations



A,B,C,D are end points of rhombus, and S is point where diagonales AC and BD meet.

My goal is to get this 2 equations:
[tex]\vec{AS}[/tex]=[tex]\stackrel{1}{2}[/tex]*[tex]\vec{AC}[/tex]
and
[tex]\vec{BS}[/tex]=[tex]\stackrel{1}{2}[/tex]*[tex]\vec{BD}[/tex]

condition of rhomb:

[tex]\vec{AB}[/tex]=[tex]\vec{DC}[/tex]
[tex]\vec{AD}[/tex]=[tex]\vec{BC}[/tex]

[tex]\vec{AB}[/tex]+[tex]\vec{BC}[/tex]+[tex]\vec{CD}[/tex]+[tex]\vec{DA}[/tex]=[tex]\vec{0}[/tex]



The Attempt at a Solution



[tex]\vec{AS}[/tex]+[tex]\vec{SD}[/tex]+[tex]\vec{DA}[/tex]=[tex]\vec{0}[/tex]
[tex]\vec{AC}[/tex]+[tex]\vec{CD}[/tex]+[tex]\vec{DA}[/tex]=[tex]\vec{0}[/tex]
_______________________________
[tex]\vec{AC}[/tex]+[tex]\vec{CD}[/tex]-[tex]\vec{AS}[/tex]-[tex]\vec{SD}[/tex]=[tex]\vec{0}[/tex]
[tex]\vec{AC}[/tex]-[tex]\vec{AS}[/tex]+[tex]\vec{BA}[/tex]-[tex]\vec{SD}[/tex]=[tex]\vec{0}[/tex]
[tex]\vec{AC}[/tex]+[tex]\vec{SA}[/tex]+[tex]\vec{DS}[/tex]+[tex]\vec{BA}[/tex]=[tex]\vec{0}[/tex]
[tex]\vec{AC}[/tex]+[tex]\vec{DS}[/tex]+[tex]\vec{SA}[/tex]+[tex]\vec{BA}[/tex]=[tex]\vec{0}[/tex]

and I don't know if I am on the right track and I wan't your opinion.
 

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  • #2
Try this. Since S is on both diagonals, you can write AS=tAC, SC=(1-t)AC, BS=sBD, SD=(1-s)BD. Now put those into AS+SD=BS+SC and AS+SB=DS+SC. Now find s and t.
 
  • #3
Dick said:
Try this. Since S is on both diagonals, you can write AS=tAC, SC=(1-t)AC, BS=sBD, SD=(1-s)BD. Now put those into AS+SD=BS+SC and AS+SB=DS+SC. Now find s and t.

I only used this equation AS+SD=BS+SC. I put these formulas into it: AS=tAC, SC=(1-t)AC, BS=sBD, SD=(1-s)BD and I got this:

2tAC - 2sBD + BD - AC = 0
2tAC - 2sBD = AC - BD
Now in order equation to be correct it must be

2t = 1 and -2s = -1 (is this correct way to do this?)

From there we get s=t=1/2.
I think that is the solution.

I just don't know this thing:
Why do you need 2 formulas(AS+SD=BS+SC and AS+SB=DS+SC) when I got solution from just one?
 
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  • #4
I 'needed' two equations because that's the first thing that popped into my head and I didn't look for a shorter route. Well done.
 
  • #5
So it is correct. I couldn't have done it without your help.Thank you very much for your help.
 
  • #6
borovecm said:
So it is correct. I couldn't have done it without your help.Thank you very much for your help.

Actually, thinking it over. Your proof is only valid if AC and BD are linearly independent. That's what allows you to equate the coefficients. If you know what 'linearly independent' means and are happy with that, ok. Otherwise, you may want to go the longer route.
 
  • #7
I try to it but I can't get to the solution. I did this "AS=tAC, SC=(1-t)AC, BS=sBD, SD=(1-s)BD. Now put those into AS+SD=BS+SC and AS+SB=DS+SC." I get these 2 equations:

tAC - sBD = (s-1)BD + (1-t)AC
tAC +(1-s)BD = sBD + (1-t)AC
_________________________
then I multiply second equation with (-1)
tAC - sBD = (s-1)BD + (1-t)AC
-tAC +(s-1)BD = -sBD -(1-t)AC
__________________________
now we sum them
tAC - tAC -sBD + sBD + (s-1)BD - (s-1)BD + (1-t)AC - (1-t)AC = 0
that totals to
0 = 0

How did you get solution??
 
  • #8
It appears that I got the solution by making a sign mistake. Sorry, I'm making a mess of this. The problem is coming since if AC is parallel to BD (and the parallelogram flattens into a line), then you can't prove much about s and t (since the diagonals intersect in lots of points). So I think your previous approach is the right one. Which means you do need some notion of linear independence. Do you know, for example, if AB and CD are not parallel, then the only solution to s*AB+t*CD=0 is s=0 and t=0??
 
  • #9
Dick said:
It appears that I got the solution by making a sign mistake. Sorry, I'm making a mess of this. The problem is coming since if AC is parallel to BD (and the parallelogram flattens into a line), then you can't prove much about s and t (since the diagonals intersect in lots of points). So I think your previous approach is the right one. Which means you do need some notion of linear independence. Do you know, for example, if AB and CD are not parallel, then the only solution to s*AB+t*CD=0 is s=0 and t=0??

This seems logic to me. What happens if AB and CD are parallel? What are the solutions then? What kind of notion do I have to put in my problem so it will be correct?
 
  • #10
If AB and CD are parallel, then AB=r*CD for some r so the eqn becomes s*r*CD+t*CD=0 or s*r+t=0 or s=-t/r. For every t there is a corresponding solution for s. What you have to put into the problem is that (in terms of the vectors in your problem and as you said) that 2tAC - 2sBD = AC - BD -> (2t-1)AC=(2s-1)BD. Now since AC and BD are not parallel, the only solution is (2t-1)=0 and (2s-1)=0. I don't know if there is an explicit statement of that in your textbook. But this problem does need something like that.
 
  • #11
I still don't get it. Is there any other way in proving this? I just have to put this ("If AB and CD are parallel, then AB=r*CD for some r so the eqn becomes s*r*CD+t*CD=0 or s*r+t=0 or s=-t/r. For every t there is a corresponding solution for s.") in my textbook?
 
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  • #12
Ok, try this. The midpoint of B and D is S=(B+D)/2. Since BS=S-B=(B+D)/2-B=(D-B)/2=BD/2. The midpoint of A and C is T=(A+C)/2. Now you just have to prove S and T are the same point. Subtract them and see if you can prove the difference is zero. This sidesteps the issue of whether the intersection is unique.
 
  • #13
Thank you for your help. I had math today. Teacher said it is okay. I supposed that vectors AC and BD are not parallel. I got one point in math with this question.
 

1. How do you prove that the diagonals of a rhombus split in a ratio of 1/2?

To prove that the diagonals of a rhombus split in a ratio of 1/2, we can use the properties of a rhombus and basic geometry. Specifically, we can use the fact that a rhombus has opposite sides that are parallel and equal in length, and that its diagonals bisect each other.

2. What is the formula for finding the ratio of the diagonals in a rhombus?

The formula for finding the ratio of the diagonals in a rhombus is as follows: the ratio of the diagonals is equal to the ratio of the lengths of any two adjacent sides. In other words, if we label the lengths of the diagonals as d1 and d2, and the lengths of the adjacent sides as a and b, then the ratio d1/d2 = a/b.

3. Can you provide a visual representation of how the diagonals of a rhombus split in a ratio of 1/2?

Yes, we can provide a visual representation of this concept. If we draw a rhombus and label its diagonals as d1 and d2 and its adjacent sides as a and b, we can see that d1 is equal to the sum of a and b, while d2 is equal to the difference between a and b. This means that d1/d2 = (a+b)/(a-b). When simplified, this ratio becomes 1/2.

4. Are there any other ways to prove that the diagonals of a rhombus split in a ratio of 1/2?

Yes, there are other ways to prove this concept. One way is to use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. By creating a right triangle with one of the diagonals of the rhombus as the hypotenuse, we can use this theorem to prove that the diagonals split in a ratio of 1/2.

5. How is knowing the ratio of the diagonals in a rhombus useful?

Knowing the ratio of the diagonals in a rhombus can be useful in many applications, such as in engineering, architecture, and construction. It can also be used in solving various geometrical problems and in understanding the properties of rhombuses and other geometric shapes. Additionally, this concept is important in the study of vectors and their properties in mathematics and physics.

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