Complete the following proof.

prove: in an equilateral triangle the three medians are equal.

(fill in the blanks of the equation in the second picture with the correct number/letter/sign based off the first picture.)

Tell you what. I'll find the length of the medians and you can fill in the blanks where they belong.
BR
B = (2a,0)
R = (a/2,b/2) Remember these things are medians. They go to the 1/2 way point of the line opposite the vertex they face.
BR^2 =  (2a - a/2)^2 + (0 - b/2)^2
BR^2 = (3/2 a) ^2 + b^2 / 4
BR^2 = 9/4 a^2 + b^2 / 4  We need to find some relationship between a and b.
Let's try AB = BC  
AB = 2a
BC = (2a - a)^2 + (b - 0)^2
BC = sqrt(a^2 + b^2)
AB = BC
2a = sqrt(a^2 + b^2) Square both sides.
4a^2 = a^2 + b^2 Subtract a^2 from both sides.
sqrt(3a^2) = sqrt(b^2)
sqrt(3)a = b

Let's put b into BR^2
BR^2 = 9/4 a^2 + 3B^2 / 4
BR^2 = 12 a^2 / 4
BR^2 = 3a^2 
BR = sqrt(3) * a

CP
C = (a,b) ; P = (a,0) This is another application of the distance formula, and it is a good one.

CP^2 = (a - a)^2 + (b - 0)^2
CP^2 = 0 + b^2
CP^2 = b^2 which you can see from the diagram. 
CP = sqrt(b^2)
CP = b but b = sqrt(3) * a 
CP = sqrt(3)*a

AQ
A = (0,0)
Q = (3/2) a, b/2
AQ^2 = (3a/2 -0)^2 + (b/2 - 0)^2
AQ^2 = 9a^2/4 + b^2/4
but b = sqrt(3) * a 
AQ^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4
AQ^2 = 9 a^2/ 4 + 3 a^2 / 4
AQ^2 = 12 a^2/4
AQ^2 = 3 a^2
AQ = (sqrt 3) * a which agrees with the other two.