# Find Length of Median in Triangle via Cosine Law

• Umaxo
In summary, the conversation discusses a question about deriving the length of a median of a triangle using cosine law. The person asking the question is getting incorrect results and is seeking help to identify their mistake. The moderator suggests checking out Apollonius's theorem and noting that the median does not bisect the vertex angle in half.

#### Umaxo

<Moderator's note: Moved from a technical forum and thus no template.>

Hi,

i am quite embarrassed to ask this question, but i am really stuck.

i want to derive length of median of triangle from cosine law and i am getting wrong results. I cannot spot the mistake.

So let's have a triange with sides a,b and c. I would like to find length of a median from vertex A to its opposite side a. Let's call the legth d. So i get two triangles from my original one. One has sides a/2,b,d and the second a/2,c,d. Thus i can write two cosine laws:

$$a^2/4=b^2+d^2-2*b*d*\cos(\alpha/2)$$
$$a^2/4=c^2+d^2-2*c*d*\cos(\alpha/2)$$
where ##\alpha## is angle in vertex A. When i get rid of cosine and solve for d, i get:
$$d=\sqrt (b*c+a^2/4)$$

which is wrong (the answer should look like this https://en.wikipedia.org/wiki/Median_(geometry), or one can just try the formula for equilateral triangle (in this case both equations reduce to the same, but i guess one can take the limit to be able to apply formula even for this case)). Can you help me to spot the mistake?
Thanks:)

From your Wikipedia ref, follow the link to Apollonius's theorem and check out the proof. Note that m=a/2.

Okay, i realized median doesn't disect vertex angle in half. Silly me:)