# Thread: Angle bisection problem, with very few measures

1. ## Re: Angle bisection problem, with very few measures

Originally Posted by bjhopper
Hi Zellator,
If I understand your new post you want to know how to find the lengths of AB and AC using Stewards Theorem. I am not familar with it but I have a plan for you.
Draw a vertical line length 6 Mark upper end A and lower D.At point D swing two arcs lengths of 3 and 4.A straight line thru D defines two points B and C.Note that you can draw many triangles.If you fix the length of either AB or AC the other line can be calculated using the cosine law

I understand most of your explanation but I didn't understand some parts, where you said to swing arcs and a vertical line.
If you ever wish to make some quick graphs, not only on this thread. There's a tool called CaRMetal that is by java, so you don't have to install anything.
So we can better understand you. Search CaRMetal on Google, it is a French program

What do you mean with many triangles?
And what do you mean with fixing the length?

Here's what I got
$\displaystyle 9=36+x^2-12xcos{(a)}$
$\displaystyle cos{(a)}=\frac{+x^2+25}{12x}$
and
$\displaystyle 16=36+y^2-12ycos{(a)}$
$\displaystyle cos{(a)}=\frac{+y^2+20}{12y}$

If we were to equal those two equations, would we get anywhere?

the theorem is $\displaystyle a(mn+d^2)=b^2m+c^2n$
But we can't use it here with that few measures.
But I didn't want to do it with it,

The last time I did this question, I didn't get it right. That's why I am asking for it again.
I used a property of right triangles, and we don't know if this is a right triangle (eventually we do, but).
I don't know how you guys got AC=6 with the bisector theorem

If I get something wrong, it is maybe because it is late here and I am kind of tired hahaha
Thanks bjhopper!

2. ## Re: Angle bisection problem, with very few measures

Hi Zellator,
No one calculated AD you supplied it.In order to calculate one of the other sides requires a length of one of them.Given 5 for AC you can find AB using the cosine law .Try it your way. (Stewards Theorem)The angle bisector theorem does not enter the problem because the angles at A will not be equal

bjh

3. ## Re: Angle bisection problem, with very few measures

Why the angles will not be equal at A? Isn't this bisection?

We know only CD BD and AD.
If we have AC it is easy to calculate AB, but how do we calculate AC?
I don't think you understood my question

We are out of one length here to find the other
We need b or c to calculate either by Cosine's Law or by Stewart's Theorem.

By Stewart's Theorem we have
$\displaystyle 8(12+6^2)=3b^2+4c^2$

And I did by Cosine's Law in my previous post.
I still don't get it bjhopper.
Thank you for having pacience with me hahahah
Thanks.

4. ## Re: Angle bisection problem, with very few measures

Zellator,
Given only AD BD BC then AC cannot be calculated.

bjh

5. ## Re: Angle bisection problem, with very few measures

Originally Posted by Zellator
In $\displaystyle \Delta ABC$, let D be a point in BC such that AD bisects angle A. Given that AD=6, BD=4, and DC=3, find AB.
This is what the problem stated, then how is it possible to solve it?

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