Man I haven't been active in a while.
Okay, here's the problem.
The median of a triangle connects a vertex of the triangle to the midpoint of the side opposite the vertex. In triangle ABC, we have AB=3 and AC=4. Side BC and the median from A to BC have the same length. What is the length of BC?
I used the Law of Cosines to try to find the answer.
I got close: basically, BC squared is the same as the cosine of angle A.
But that's where I get stuck; I can't find angle A from the given information(or maybe I'm just not looking hard enough!).
Thanks guys! - BasicMaths
We can use the Law of Cosines twice to find BC
Let the intersection of the median drawn from A to BC = D
The angles formed by the intersection of the median with BC will be supplementary...so, their cosines will have opposite signs.....so cos BDA = - cos CDA .....so -cos BDA = cos CDA
So we have that
3^2 = [ (1/2) BC]^2 + (BC)^2 - 2[ (1/2)BC] [BC] cos BDA (1)
4^2 = [ (1/2)BC]^2 + (BC)^2 - 2[(1/2)BC] [BC] cos CDA
4^2 = [ (1/2)BC]^2 + (BC)^2 - 2[ (1/2)BC][BC] (-cos BDA)
4^2 = [ (1/2)BC]^2 + (BC)^2 + 2[ (1/2)BC][BC] (cos BDA) (2)
Add (1) and (2) and we get that
25 = 2[ (1/2)BC]^2 + 2[BC]^2 simplify
25 = 2 (1/4)BC^2 + 2BC^2
25 = (1/2)BC^2 + 2BC^2
25 = (2.5)BC^2 divide both sides by 2.5
10 = BC^2
BC = √10
Here's a pic :