The three medians of the triangle are constructed with a segment from the vertex to the midpoint on the opposite side. The medians are concurrent at the orthocenter (seen below as point X) . They create 6 smaller triangles of equal area. Below is the proof that all 6 triangles have the same area.
First, see that triangle AXM has the same area as CXM. Recall that the area of a triangle is equal to (1/2)(base)(height).
base of triangle AXM = length of AM = length of MC = base of triangle CXM
height is the same on both of these triangles because it is the height from X to segment AC
Therefore, triangle AXM has the same area as CXM. Call the area p.
Next, see that triangle AXN has the same area as BXN.
base of triangle AXN = length of AN = length of NB = base of triangle BXN
height is the same on both triangles because it is the height from X to segment AB
Therefore, triangle AXN has the same area as BXN. Call the area q.
Next, see that triangle BXL has the same area as CXL.
base of triangle BXL = length of BL = length of LC = base of triangle CXL
height is the same on both triangles because it is the height from X to segment BC
Therefore, triangle BXL has the same area as CXL. Call the area r.
Now we need to see that triangle ABM has the same area as triangle CBM. Notice that each of these triangles are made up of three smaller triangles. Triangle ABM has area = p + q + q. Triangle CBM has area = p + r + r. Prove these are the same through a similar method as before:
base of triangle ABM = length of AM = length of MC = base of triangle CBM
height is the same on both triangles because it is the height from B to segment AC
Therefore, triangle ABM has the same area as triangle BXN. So.....
p + q + q = p + r + r
q = r
Finally, we can prove that the area of triangle BAL = area of triangle CAL. Keep in mind that the area of triangle BAL = q + q + r and the area of triangle CAL = p + p + r.
base of triangle BAL = length of BL = length of LB = base of triangle CL
height is the same on both triangles because it is the height from A to segment BC
Therefore, triangle BAL has the same area as triangle CAL. Using q = r, we see that.....
r + r + r = p + p + r
2r = 2p
r = p = q.
Since r = p = q, I have shown that all 6 triangles have the same area.