Can Vector Addition Solve Non-Right Triangle Proofs?

  • Thread starter Thread starter byronsakic
  • Start date Start date
  • Tags Tags
    Proof Triangle
byronsakic
Messages
17
Reaction score
0
Hello,
Here is the question:
question2.jpg

Here are 2 different ways i tried to approach the question:
In the first way, i tried using pythagoreom theorem by making a right angle, but after i did the question i realized since it did not say it was a right triangle, ic annot assume there is a 90 degrees angle and i cannot use the pythagoras theorem.
part1question2.jpg

Second way i tried using vector addition. However you can see at the end i was stuck.
part2question2.jpg

though i believe vector addition is the way to solve this problem, however we had never done any problems involving the squared of a vector unless it was using the pythagoras theorem,
any suggestions in what i did wrong or what i could do?

Sorry for my messy writing hehe.

thanks
byron
 
Physics news on Phys.org
Byron,

Try this. Drop a perpendicular from A to the line segment BC. Call its length h and the distance between the intersection point and the midpoint x. Now you can use Pythagoras to solve for x and h from which you can calculate AD.
 
I know why you can't do it with vector addition: nowhere in your work did you make use of the fact that D is the midpoint of the line segment BC.

(In fact, you didn't even made use of the fact that D is on the line through B and C!)
 
didnt i do it in the first try? but if you drop a perpendicular from A to line segment BC then how would u prove that it is a right angle? but i think i did that in first try..

though i am going to try coordinate proofs right now :D i think that is the solution
 
I only looked at your vector addition approach. I agree that ought to be the easiest solution. (unless you're particularly adept with synthetic geometry)
 
i got the answer by using coordinate proofs and making LS = RS

thanks for all your help :D
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top