HELP: Geometry Proof

Re: HELP: Geometry Proof


1> Start by showing that triangle ADE is similar to ABC.
2> Show that segment DE is one half of BC
3> Show that triangle DEF is equivilant to FCP
4> Show that segment CP is the same length as DE
5> Therefore CP is one half of BC

Quick Fix
im sorry, thanks so much for the help, but i need some clairification is that all one proof or is that several proofs in one.

Re: Quick Fix


Those are the logical steps to take in a single proof. You will have to use the theorems you have already studied (I assume) to show each one. At the end, you have a logic chain that proves what you want.

thank you
thanks, but you wouldnt happen to know how to use Vectors in geometrical proofs? i need a little push with that.

Teacher:
1. I trust completely that you know the answer to this.

2. You wouldn't be giving us this assignment if it weren't possible to prove it somehow.

Therefore, it is provable and that's good enough for me.
Q.E.D.

Re: thank you


This problem doesn't use vectors; it is a standard angles and triangles type of proof. Do you have a specific problem with vectors you are concerned with?

like external point vectors, and stuff, its just simple lines and planes vectors from gr 12 geometry not vectors from physics.

can anyone help me solving this in vector formation?

Is this the same problem?
Are you still looking for help on the initial problem? If so, I tried to help in my first response. Can you give us more details on where you're having problems? I'm not sure what you are after with "vector formation".

ok heres what i was given as an answer but i dont understand it:

Let A, B & C have position vectors a, b & c.

k & l are scalars.

OP = OB + k BC = b + k (b - c) = (1 - k) b + k c [1]

Also

OP = OD + DP = OD + l DF = OD + l (OF - OD) = 1/2(a + b) + l {3/4(c - a) -1/2(a + b)}.

This simplifies to

OP = 1/2(a + b) + l/4(3c - a - 2b) [2]

Equating equations [1] & [2] we get

l = 2 & k = 3/2 .

=> OP = 3/2 c - 1/2 b => CP = OP - OC = 3/2 c - 1/2 b - c = 1/2(c - b) = 1/2 BC .



if you can help by clarifing any of that, thank you

I see what you are after
Wow, that is a really hard way to solve a really easy problem. I'll try to explain, but a lot of this won't make sense unless you understood it in class.

Picture your triangle ABC on a Cartesian grid. We will call the origin O. Point A is at some position (Xa,Ya), B is at some position (Xb,Yb), etc. "Position vector a" means a line drawn from the origin to A. It could also be written as OA. If you wanted to get to B, you could go directly to B (position vector b, OB) or you could first drive to A down "position vector a" then to B using "position vector AB". This means that OB=OA+AB. It helps to think of this like driving. You can to the grocery store directly, or by adding trips to the gas station, gas station to video store and video store to the grocery store. This is not simple, but if you read it several times and don't get it, you need to see your teacher.

Now let's look at your proof...

OP = OB + k BC = b + k (b - c) = (1 - k) b + k c [1]

This is saying that you can get to point P by driving to point B and following the path to BC. How far you go down that path is determined by k. If k is 2, you would drive twice the distance required to get to C. Next, there is some substitution. b is just another way to write OB, so that is straightforward. Remember that you can get to C by driving to B then going to C so OC = OB + BC or c = b + BC. This means that BC = b-c. That is where the second equality comes from. The third equality is even more tricky. Since k is an unknown constant, you can multiply it by a constant and it is still an unknown constant. That is how the signs get changed. You would think that the last term should be (1+k)b - kc, but the k was replaced with -k. I'm certain the [1] on that equation explains what was done with k.

OP = OD + DP = OD + l DF = OD + l (OF - OD) = 1/2(a + b) + l {3/4(c - a) -1/2(a + b)}.

This line is similar. You can get to P by going to D and then following the path between D and P. The path from D to P is along the path from D to F, so DP can be stated as l DF just like we used k above. DF can be replaced with OF - OD like we did with BC above. Now we know that D is half way between A and B, so you can replace OD with 1/2(OA + OB) = 1/2(a+b). In order to see this, you have to think of the grid. If A is at (Xa,Ya) on the grid and B is at (Xb,Yb), you can see that the point half way between them is at the point that is half way between the X points and halfway between the Y points. That point is at the average of the X and Y coordinates, or [ (Xa+Xb)/2 , (Ya+Yb)/2 ]. Once again, pull out the graph paper and try it for yourself a few times to see that it makes sense. OF is replaced in a similar manner. Point F is 3/4 of the way between A and C.

The rest is algebra. I hope this helps, but I fear that I cannot adequately explain this without you sitting beside me drawing on graph paper. Good luck.

Thank you
well actually i just figured it out 5 minutes before i explaination, which was alot better than my understanding.

you wouldnt happen to know a better way to solve it would you?

My first reply showed how I would solve it using equivalent triangles. Triangle DEF and FCD are equivalent and ADE is similar to ABC so you know that DE is 1/2 BC. Since DE = CD, CD = 1/2 BC.

Try this
Extend DF in the opposit direction as well to a point Q such that BQ is parallel to AYou have two proportional triangles. BQ = AF = .75 AC
triangle QBP is proportional to triangle FCP

FC = .25 AC soooo.
FC : BQ = CP : BP = 1 : 3

3CP = BP = BC + CP => BC = 2 CP

QED

Just happened to notice your problem several years later...
Have a drawn out diagram with all the points labelled.
Let's say BC and CP are divided into the ratios 1-k and k respectively.

Using position vectors to represent BC and CP we get:
BC = OC - OB , CP = OP - OC

Both OC can be rewritten as:
OC = k(OB) + (1-k)OP
OC = kOB + OP - kOP or kOB + (1-k)OP
Now stick that back into the vector equation BC and CP we get:

BC = kOB + (1-k)OP - OB , CP = OP - kOB + (k-1)OP

BC = (k-1)OB + (1-k)OP , CP = -kOB + kOP
The ratio that divides BC and CP are the same so you compare the coefficents.
k-1 = -k
2k = 1
k = 1/2
Therefore 1 - k = 1/2 as well.