Problem with vectors

[urza]

hey there

can anyone help me with this math problem concerning vectors:

 

Given a triangle ABC. Let D and E be the points defined by BD= 2/3 BC and AE=1/2 AC. The lines (BE) and (AD) intersect at M, and the line thorugh E parallel to (BC) intersects (AD) in F.

 

a) Prove that MD + 4MF = 0 and MB + 4ME = 0. Deduce that 2MA + 3MD=0

b) Show that 5MC = 2CA + CB

c) Define L by CL + 2 CA = 0 and I by LI = BC. Prove that C, M, and I are

collinear.

 

thx

[5614]

It can sometimes help by drawing out the problem so that you can visualise it... once you can see it in front of you solutions can often see more obvious.

[Severian]

Am I the only one who gets pissed off with people coming here and asking us to do their homework for them? This isn't supposed to be a forum for us to do their homework - they are supposed to ask questions and explain where they are stuck so that we can guide them towards better understanding.

[Dave]

Quite

 

Urza, we don't do the work for you. You're going to have to try it first, and if you get stuck then we'd be willing to offer pointers.

[reverse]

hey there

can anyone help me

 

The trick here is to be aware of the rules about triangles you already know.

 

draw it out as good as you can.

 

then check it.

 

then look at it for ages and see how many ways you can apply the simple rules you already know.

 

usually the answer is in the question itself.

 

to me the words that stick out further that the spikes of a cactus are the following.

 

triangle

intersect

parallel

collinear.

 

 

 

good luck.

[urza]

well obviously i wouldn't be here if i wasnt stuck

and no this isnt a homework assignment

and yes i've already drawn the triange in order to visualize it

and you can't do it using simple triangle rules...

 

plus have any of you tried doing it before you answered here

thx anyway..

[reverse]

how were you going to start?

 

 

(I'm well rusty on this stuff).

 

Why do you suppose the first thing asked is to show how three equations all come to equal "0".

 

does that sound like anything you have heard before?

[mezarashi]

I feel sorry for urza, so I thought I'd give it a try since I've not much better to do

 

"and you can't do it using simple triangle rules..." - urza

 

This statement isn't true at all. Most of these geometry problems require a very good understanding of the basic rules. This problem turns out to be quite challenging and I did enjoy thinking about it for a bit. Very tedious nonetheless, and I'll leave the 2 later parts for you to try out. It is solved using "basic" intuitive rules, but if you feel insecure, you can try adding in some of the trigonometry if you are math savvy (e.g. sine rule) and you will see that the two approaches yield the same answer.

 

 

(a)i. Prove that MD + 4MF = 0

 

This is the same as saying that MD = 4FM, since FM = -MF. They are collinear so all we have to do is prove that MD is 4 times the length of FM. For the rest of my explanation, I will not take the direction into account. So I'll say things like MD = 4FM as well as MD = 4MF interchangingly. I've drawn a diagram of an arbitrary triangle. Note the congruency indicated by the colors due to the parallel line QE.

 

 

In addition, you will notice that triangle EFM is congruent with triangle BDM. This can be verified by showing that all the internal angles are the same.

 

<FME = <DMB : this is due to the reflection rule of two intersecting lines

<AFQ = <ADB (<MDB) : Congruent triangles (green) because of the parallel line.

<AFQ = <EFM : reflection of two intersection lines again

<FEM = 180 - <EFM - <FME : Sum of triangle's internal angles always = 180

 

But we know that <EFM = <MDB and <FME = <DMB, so <FEM = DBM.

 

 

Having established the congruency of EFM and BDM, we can then move on to to proving that MF (a side of the smaller triangle) is equal to 1/4 MD. To do this we simply need to prove that ANY side of the smaller congruent triangle is 1/4 the length of the corresponding side in the larger congruent triangle.

 

My approach was with using the blue congruent triangles (ABC) and (AQE).

Since AE = 1/2 AC, then by congruency, QF = 1/2 BD and most importantly FE = 1/2 DC.

 

But we know that DC = 1/3 BC. We also know that BD = 2/3 BC = 4/6 BC.

With that, FE = 1/6 BC.

 

So we see here that the corresponding sides FE = 1/4 BD, as FE = 1/6 BC and BD = 4/6 BC.

 

Since FE = 1/4 BD, then ME = 1/4 MB, and ultimately MF = 1/4 MD. If you don't trust your instincts here brought to you by the congruency rule, then apply the sine rule.

 

This completes the explanation for 4MF = MD.

 

 

(a)ii. Show that MB + 4ME = 0

 

This was shown in part (i) " Since FE = 1/4 BD, then ME = 1/4 MB, and ultimately MF = 1/4 MD. "

 

(a)iii. Deduce that 2MA + 3MD=0 or that 2MA = 3DM

 

This requires just a bit of arithmetic, no trig involved (I suck at trig). Equations we have from the diagram.

 

AD = MA + MD

AF = 1/2 AD

AF + MF + MD = AD; MF + MD = AF = 1/2 AD

 

From part (i), we know now that MD = 4MF

 

(right hand side)

3DM = 3(4MF) = 12MF

 

(left hand side)

2MA = 2(DA - DM)

= 2DA - 2DM

= 2( 2(DM + MF) ) - 2DM

= 4DM + 4MF - 2DM

= 2DM + 4MF

= 2(4MF) + 4MF

= 8MF + 4MF

2MA = 12MF

 

=D

And we're done =D

 

The rest of the question relies on the same elementary principles. Keep looking for clues in the congruencies, good luck and have fun

[urza]

OMG

 

thx alot mezarashi, you're a lifesaver

 

this is more than i expected to get

 

thx alot for your time

[reverse]

check out this...

 

it should give you the idea on why the simple stuff works.

 

http://everyschool.org/u/logan/culturalmath/pythag.htm