# God how do you prove geometry theorems?

I would really appreciate it if some1 can help me... We are doing basic vectors about adding, multiplying by scalars, subtracting etc...I have a problem with the following question and something else in general:

1)I attached an image and I have to determine what vectors AB-BC+CD-DE+EF-FA will equal

IMAGE AT: http://f1.pg.briefcase.yahoo.com/bc/robvdamm/lst?.dir=/Math [Broken]

Click on "vector"

2)How can I prove a theorem?

What I did for number one is that I drew another point in the middle and called it G (Its not in the picture so you have to imagine there is a point in the middle) and I named vectors GC parallel to AB. BUT my teacher says prove that, how the hell can I prove it? I would kill geometry if I understood how to prove the stuff...Any help is much appreciated, thanks!

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dduardo
Staff Emeritus
Just by looking at the picture visually, It looks like the answer is going to be zero.

Here is the method I took:

(AB->) = -(DE->)
(CD->) = -(FA->)
(BC->) = -(EF->)

Right off the bat I can elimate 3 terms

(AB->) - (BC->) + (CD->) - (DE->) + (EF->) - (FA->)

becomes...

2 * [ (EF->) - (FA->) - (DE->) ]

Then by breaking up each vector into its i^ (x) and j^ (y) components I cancel EFj^ and FAj^ which leaves me with EFi^ + FAi^. But since the picture is of a regular hexagon EFi^ = FAi^

so I can say 2* [ 2*EFi^ - (DE->) ]

Again, knowing that it is a regual hexagon, 2*EFi^ - (DE->) = 0

You can prove this by using the interior angles (120 degrees) and pluging in some value for the length of the hexagon and calculate my solution.

HallsofIvy
Homework Helper
I would think (1) would be kind of obvious. Well, almost obvious. Your picture is of a regular hexagon (I assume that the length of all vectors is the same. That's what it looks like. You don't specifically say that. Was it given that the lengths are all equal?) with the vectors all going "counter-clockwise". Did you notice that to every vector there is an "equal and opposite" vector- the side opposite it? If the problem were AB+ BC+ CD+ DE+ EF+ FA, then it really would be obvious! Either because the "equal and opposite" vectors cancel or simply because the "arrows" go around in a circle and end exactly where they started.

Since the problem is AB- BC+ CD- DE+ EF- FA, we can note that -DE=
AB, -BC= EF and -FA= CD so this the same as 2(AB+ CD+ EF). Well, I'll be darned! Those three vectors form an equilateral triangle: that sum is 0 too!

There is no "Royal Road" to proving statements. As far as your "GC" is concerned, I would look at the picture and suspect that the four points A, B, C, D form a parallelogram. If I could prove that was a parallelogram, then, of course, opposite side are congruent. If I remember my geometry correctly, then one way to prove "parallelgram" is to prove one pair of opposite sides is both
parallel and congruent. How would I prove that? Well, since this is a regular hexagon, what do I know about the angles? Hmm, If I were to draw a line from the center (G) to EVERY vertex, that would divide the center into 6 congruent angles: 360/6= 60. That seems awfully familiar. Aha! equilateral triangles! That divides the hexagon into 6 equilateral triangles! That at least proves that side AG is congruent to side BC. Now look at the angles: Angle GAB is, of course, 60 degrees. If I extended AB past B what would that angle be?
The three angles, ABG, CBG and this (unnamed) angle form a straight line so add to 180 degrees. ABG and CBG are both 60 degrees (from above) and so add to 120 degrees that leaves, for that external angle,
180- 120= 60 degrees! OK!! That's the same as angle GAB! Remember "if corresponding angles on a transversal are congruent then the lines are parallel"? Now that I know AG and BC are both congruent and parallel it follows that ABCG is a parallelogram and thus CG is parallel to AB.

Do you see what I did? I looked at the picture and I saw a parallelogram. That started me thinking about every thing I knew about parallelograms and shifted the problem from "prove AB and GC are parallel" to "prove ABCG is a parallelogram". Once I was thinking of that, I thought about opposite sides being congruent and parallel and then to angles. You have to think "What would have to be true so that one of the theorems I know will give me the result I want? What do I have to know to prove THAT is true?" and just keep working back until you get to something you are given.

By the way, the way I did (1), I asserted that "Those three vectors form an equilateral triangle". I'm sure your teacher would insist that I prove that too! I would do that just the way I did above: Draw lines from each vertex to it's opposite vertex, dividing the hexagon into equilateral triangles. I would then use the fact that the angles are all 60 degrees to show that the 3 vectors AB, CD, and EF, moved together, form an equilateral triangle.

dduardo
Staff Emeritus
a little late hitting submit Hallsoflvy

I like vector notation better

Thanks for the replies guys. Yes you both say that AB=-DE and BC=-EF etc... I told my teacher the same thing and she says to prove that, I told her its obvious but she says that I cant just do that I have to somehow make it more obvious to her. Also , dduardo, "i^ (x) and j^ (y)"

dduardo
Staff Emeritus
i^ or eye hat is the x unit vector
j^ or jay hat is the y unit vector

I put x and y in parathesis just to make it clear I was dealing with the x and y components

I don't know if your familiar with this type of notation, but each vector can be broken down into their spatial components.

a vector (v->) in two dimensions can be broken down into x and y components

|v->| i^ = |v->| cos( theta )
|v->| j^ = |v->| sin( theta )

where |v->| is the magnitude of the vector
and theta is the angle of the vector

Just imagine a right triangle where you know the hypotenuse (vector magnitude) and you want to figure out the two legs.

HallsofIvy