Use vector methods to prove median from A is perp. to BC

[Slash93]

Im not sure if vectors fit in this category but any help would be appreciated. Heres my question-

ABC is an isoceles triangle with base BC. Use vector methods to prove the median from A is perpendicular to BC.

Im almost certain I am to use a dot product but I could be wrong. Any little push to get me started would be great.
Thanks!

 

[Deleted member 4993]

Re: I need a start

Hey!
Im not sure if vectors fit in this category but any help would be appreciated.
Heres my question-

ABC is an isoceles triangle with base BC. Use vector methods to prove the median from A is perpendicular to BC.

Im almost certain I am to use a dot product but I could be wrong. Any little push to get me started would be great.
Thanks!

Locate point A at (a,0) and B at (b,0) locate point C at (b,c)

Find |AC|^2 and |BC|^2 and equate those (isoscales condition) ..(1)

Locate the point of intersection (D) of median on AB <----- definition of median

Now define CD and AB as vectors and use dot product condition.

Please show your work - so that we can help you better.

 

[Slash93]

I dont understand how you located those points the way you did? The way i drew the triangle was with the base BC on the x-axis with B(-1,0) or (b,0) and C (1,0) or (c,0)?

BC is the base, not AB.

 

[Deleted member 4993]

I dont understand how you located those points the way you did? The way i drew the triangle was with the base BC on the x-axis with B(-1,0) or (b,0) and C (1,0) or (c,0)?

You can do that!

I had AC = BC and You have AB = AC

 

[Slash93]

Okay. If BC was on the x-axis and A was along the y-axis, would these points work:
A(0,a)
B(b,0) (this may be negative b)
C(c,0)
and D would be the origin

I could really used some major help!!!
Pleaseee

 

[pka]

ABC is an isosceles triangle with base BC. Use vector methods to prove the median from A is perpendicular to BC.

I would do with purely vector methods.
We are given that \(\displaystyle \vec {AB} \,\& \,\vec {AC}\) are two non-parallel vectors of the same length.
Suppose that D is the midpoint of \(\displaystyle \vec {BC}\).
It is easy to see that \(\displaystyle \vec {AD} = \vec {AB} + \vec {BD} = \vec {AC} + \vec {CD}\).
Because the base angles are congruent we have that \(\displaystyle \vec {AB} \cdot \vec {BD} = \vec {AC} \cdot \vec {CD}\).
Can you use these to show that \(\displaystyle \vec {BC} \cdot \vec {AD} = 0\)

 

[Slash93]

hmm... I see what you are getting at, I just dont understand how i can bring in the BC to show that BC (dot) AD = 0 without using components?

 

[McLovin']

I don't understand how BC dot AD=0 and how this will prove that AD is perpedicular to BC.

 

[pka]

I don't understand how BC dot AD=0 and how this will prove that AD is perpedicular to BC.

We know that \(\displaystyle \left\| {\vec {AB} } \right\| = \left\| {\vec {AC} } \right\|\quad \& \quad \left\| {\vec {BD} } \right\| = \left\| {\vec {CD} } \right\|\) and
\(\displaystyle \vec {AD} + \vec {DB} = \vec {AB} \quad \Rightarrow \quad \left\| {\vec {AB} } \right\|^2 = \left\| {\vec {AD} } \right\|^2 + 2\vec {AD} \cdot \vec {DB} + \left\| {\vec {DB} } \right\|^2\).
Likewise, \(\displaystyle \vec {AD} + \vec {DC} = \vec {AC} \quad \Rightarrow \quad \left\| {\vec {AC} } \right\|^2 = \left\| {\vec {AD} } \right\|^2 + 2\vec {AD} \cdot \vec {DC} + \left\| {\vec {DC} } \right\|^2\)

Combining the two we get:
\(\displaystyle \begin{array}{l}
\left\| {\vec {AD} } \right\|^2 + 2\vec {AD} \cdot \vec {DB} + \left\| {\vec {DB} } \right\|^2 = \left\| {\vec {AD} } \right\|^2 + 2\vec {AD} \cdot \vec {DC} + \left\| {\vec {DC} } \right\|^2 \\
2\vec {AD} \cdot \vec {DB} = 2\vec {AD} \cdot \vec {DC} = - 2\vec {AD} \cdot \vec {DB} \\
4\vec {AD} \cdot \vec {DB} = 0\quad \Rightarrow \quad \vec {AD} \bot \vec {DB} \\
\end{array}\)

 

[Slash93]

hmm

I understand all that except for the part where you stated...

2AD (dot) DB = 2AD (dot) DC = -2 AD (dot) DB

How did you come to this?

 

[stapel]

I don't understand how BC dot AD=0 and how this will prove that AD is perpedicular to BC.

It gets very confusing when you "reply" from a different one of your accounts. Please pick one of "Slash93", "Knight93", and "McLovin'" for any given thread, and then use that same personality throughout.

Thank you.

Eliz.

 

[pka]

Re: hmm

2AD (dot) DB = 2AD (dot) DC = -2 AD (dot) DB
How did you come to this?

\(\displaystyle \vec {BD} = - \vec {CD}\)

 

[Slash93]

Im not Mclovin ( i think thats my buddy from school), but I am Knight93. I just forgot my password to that account so i had to ake this one. Sorry.

oooo i beleive i see it now. Thanks for your help Pka!