# Midpoint Theorem

• September 26th 2012, 11:19 PM
powercroat783
Midpoint Theorem
A triangle has vertices A (a,b) , B (c,d) and C (e,f). Prove two things - algebra only - that the line segment connecting the midpoint of two sides of the triangle is parallel to the third side, and ​the line segment is also equal to one-half the length of the third (parallel) side.
• September 26th 2012, 11:22 PM
MarkFL
Re: Midpoint Theorem
Do you know how to find the coordinates of the mid-point of a side of the triangle?
• September 26th 2012, 11:38 PM
powercroat783
Re: Midpoint Theorem
It would be just using the midpoint formula

So let's say the midpoint of A and B is ( (a+c)/2 , (b+d)/2 ) and the midpoint of A and C is ( (a+e)/2 , (b+f)/2 ).

I believe.
• September 26th 2012, 11:52 PM
MarkFL
Re: Midpoint Theorem
Yes, that's correct. Now, how can we show that two line segments are parallel?
• September 26th 2012, 11:57 PM
powercroat783
Re: Midpoint Theorem
Well let me call AB's midpoint D and AC's midpoint E.

Having a line go through BC and DE and producing similar angles would prove this to be true... and convenient since AB and AC do happen to this. But i'm supposed to prove it algebraically, which is where I am stuck.
• September 27th 2012, 12:08 AM
MarkFL
Re: Midpoint Theorem
Hint: Two lines are parallel if their slopes are equal...
• September 27th 2012, 12:27 AM
powercroat783
Re: Midpoint Theorem
Well the Slope of the line DE is change in y / change in x so: ( (b+d)/2 - (b+f)/2 ) / ( (a+c)/2 - (a+e)/2 ) )

Simplify once i get ( (b+d-b-f/2) / (a+c-a-e/2) )

Simplify again i get ( (d-f/2) / (c-e/2) )

and again (d-f/c-e)

Slope of the line BC is ( (d-f) / (c-e) )

Well isn't math just magic :D

So all that leaves is to prove the line segment DE is half the size of BC, and I am sure I can accomplish that with the distance formula?
• September 27th 2012, 12:31 AM
MarkFL
Re: Midpoint Theorem
Yes, the distance formula will work wonders for you! :)
• September 27th 2012, 12:48 AM
powercroat783
Re: Midpoint Theorem
So the distance formula for DE is sqt { [(a+c)/2 - (a+e)/2]2 + [(b+d)/2 - (b+f)/2]2 }

So sqt {
(c-e/2)2 + (d-f/2)2 }

And distance formula for BC is sqt { (c-e)2 + (d-f)2 }

But that doesn't look like definite proof to me ( I could be wrong ), is it possible to simplify this further?
• September 27th 2012, 01:05 AM
MarkFL
Re: Midpoint Theorem
Yes, you may write:

$\sqrt{\left(\frac{c-e}{2} \right)^2+\left(\frac{d-f}{2} \right)^2}=\sqrt{\left(\frac{1}{2} \right)^2((c-e)^2+(d-f)^2)}=$

$\frac{1}{2}\sqrt{(c-e)^2+(d-f)^2)}$
• September 27th 2012, 01:13 AM
powercroat783
Re: Midpoint Theorem
Oh wow. Thanks again. So just one quick question, how do you display your math like that, with the fancy text? I was thinking that hopefully I would be able to help people at lower levels of math rather than just leach off the forum for answers. Hopefully I can contribute more and knowing how to do that would help plenty.
• September 27th 2012, 01:30 AM
MarkFL
Re: Midpoint Theorem
It is accomplished using $\LaTeX$. Teaching its usage is really beyond the scope of a single post, but if you do a search here and online for using latex, you will find lots of good information.

One technique I used when I was learning to use it is to quote the posts of others and look at the code they use without submitting the quoted post. ;)