Find Length of Median in Triangle via Cosine Law

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To derive the length of a median in a triangle using the cosine law, the user attempts to apply the law to two resulting triangles formed by the median. They initially set up equations based on the cosine law but arrive at an incorrect formula for the median length. The correct formula for the median can be found through Apollonius's theorem, which clarifies that the median does not bisect the vertex angle. The user acknowledges their mistake regarding the angle bisector property. The discussion emphasizes the importance of correctly applying geometric principles when deriving formulas.
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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:)
 
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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:)
 
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