Find the midpoint of a triangle with three endpoint given

**DBA** · November 26th 2010, 01:11 PM

Hello,
I try to figure out a formula to find the mid point of a triangle.
Given are the 3 endpoints A(x,y), B(x,y) and C(x,y).

I can find the side length a, b,c of the triangle by using the distance formula.

$d = \sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2}$

Then I could find the angles with the Law of Cosine.

$A = \cos^{-1}(\frac{b^2+c^2-a^2}{2bc})$

$B = \cos^{-1}(\frac{a^2+c^2-b^2}{2ac})$

$C = \cos^{-1}(\frac{a^2+b^2-c^2}{2ab})$

The intersection, m(x,y), of the angle bisectors is the center of the incircle.
But how can I find this intersection with the information I have?

Thanks for any help!

**Plato** · November 26th 2010, 01:23 PM

If we can use vectors then this is useful.
Let $p = \overrightarrow {AB} \;,\,q = \overrightarrow {AC}$ then the vector $r=\left\| p \right\|q + \left\| q \right\|p$ bisects the angle $\angle BAC$ .

**DBA** · November 26th 2010, 01:48 PM

Hello,
thanks for the answer but I do not understand it.
I am searching for the midpoint m(x,y).
I did not work with vectors yet. So, what exactly does p and q stands for? And how do I get the x-value and the y-value of the point?
I cannot find the mid point by drawing it because I want to compute it by a computer.

Thank you.

**Plato** · November 26th 2010, 02:16 PM

Clearly if you do not understand vectors, then that approach is useless to you.

**DBA** · November 26th 2010, 03:44 PM

Ok, thank you. I will try to read myself into vectors to understand it.

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