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

1. Nov 7, 2007

### borovecm

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

1. The problem statement, all variables and given/known data

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.

2. Relevant 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:
$$\vec{AS}$$=$$\stackrel{1}{2}$$*$$\vec{AC}$$
and
$$\vec{BS}$$=$$\stackrel{1}{2}$$*$$\vec{BD}$$

condition of rhomb:

$$\vec{AB}$$=$$\vec{DC}$$
$$\vec{AD}$$=$$\vec{BC}$$

$$\vec{AB}$$+$$\vec{BC}$$+$$\vec{CD}$$+$$\vec{DA}$$=$$\vec{0}$$

3. The attempt at a solution

$$\vec{AS}$$+$$\vec{SD}$$+$$\vec{DA}$$=$$\vec{0}$$
$$\vec{AC}$$+$$\vec{CD}$$+$$\vec{DA}$$=$$\vec{0}$$
_______________________________
$$\vec{AC}$$+$$\vec{CD}$$-$$\vec{AS}$$-$$\vec{SD}$$=$$\vec{0}$$
$$\vec{AC}$$-$$\vec{AS}$$+$$\vec{BA}$$-$$\vec{SD}$$=$$\vec{0}$$
$$\vec{AC}$$+$$\vec{SA}$$+$$\vec{DS}$$+$$\vec{BA}$$=$$\vec{0}$$
$$\vec{AC}$$+$$\vec{DS}$$+$$\vec{SA}$$+$$\vec{BA}$$=$$\vec{0}$$

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

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Last edited: Nov 7, 2007
2. Nov 7, 2007

### Dick

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. Nov 8, 2007

### borovecm

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?

Last edited: Nov 8, 2007
4. Nov 8, 2007

### Dick

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. Nov 8, 2007

### borovecm

So it is correct. I couldn't have done it without your help.Thank you very much for your help.

6. Nov 8, 2007

### Dick

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. Nov 9, 2007

### borovecm

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. Nov 9, 2007

### Dick

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. Nov 10, 2007

### borovecm

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. Nov 10, 2007

### Dick

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. Nov 12, 2007

### borovecm

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?

Last edited: Nov 12, 2007
12. Nov 12, 2007

### Dick

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. Nov 13, 2007

### borovecm

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.