Thread: 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

Hi bjhopper! Thanks again for your reply

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


and



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

About the Stewart's Theorem,
the theorem is
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!

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

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


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.

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


bjh

Originally Posted by Zellator

In , 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?