Derivation : How to find the length of a median of a triangle

[@rkk-ho3pe]

Hi, Can you let me know which whiteboard app do you use for teaching ? Also, is it available on windows ?

[@murdock5537]

Nice! ∆ ABC → BC = a = a/2 + a/2 = BD + CD; AB = c; AC = b; AD = m/2 = ?
AE = 2AD = AD + ED = m → AC = BE↔AB = CE →
m^2 + a^2 = 2(b^2 + c^2 ) → m/2 = (1/2)√(2(b^2 + c^2 ) - a^2)

[@vijayanmaster-x3r]

Nice

[@giuseppemalaguti435]

Angoli A,B,C,opposite fron a,b,c...m=(bsinC/2sinB)√(1+(b/c)^2+2bcosA/c)...gli angoli A,B,C si possono calcolare con Briggs

[@tontonbeber4555]

I still hate geometry so ...
OK lets def coord ...
B(-a/2,0)
C(a/2,0)
A(x,y)
dist AB : (x+a/2)²+y²=c² (I)
dist AC : (x-a/2)²+y²=b² (II)
(I)-(II) : 2ax = c²-b²
x = (c²-b²)/2a
m²=x²+y²
(I) : x²+ax+a²/4+y²=c²
m²= c²-a²/4 -ax = c² -a²/4 -c²/2 +b²/2 = (1/4) (2b²+2c²-a²)
answer = (1/2) sqrt (2b²+2c²-a²)