Question

For a triangle ABC with |AB| = c, |AC| = b, |BC| = a, let M be the midpoint of AB. Find |CM|

Answer 1

Say \(\angle AMC = \theta\), then \(\angle BMC = 180-\theta\) as they both are linear pair angles.

Answer 2

Apply law of cosines in the left side triangle and get
\[\large b^2 = |CM|^2+(c/2)^2-2|CM|(c/2)\cos(\theta)\tag{1}\]

Apply law of cosines in the right side most triangle and get
\[\large a^2 = |CM|^2+(c/2)^2-2|CM|(c/2)\cos(180-\theta)\tag{2}\]

Answer 3

Adding them gives the desired result but im not so sure if you're allowed to use law of cosines

Answer 4

I don't think it's not allowed. We were supposed to do it 2 ways. I'm sure something like this is fine for the first way, but for the 2nd way it had to be done with vectors.