To complete the proof that all three medians of a triangle intersect at one point, we can fill in the missing pieces in the steps as follows:
Step 1:
Draw each median so that it starts at a vertex and ends at the midpoint of the opposite side.
Step 2:
The midpoint between (0, 0) and (b, c) is ((b/2), (c/2)).
The midpoint of (0, 0) and (a, 0) is ((a/2), 0).
The midpoint of (a, 0) and (b, c) is ((a + b)/2, (c/2)).
Step 3:
Use the formula (1/3) * vertex1 + (2/3) * midpoint to find the coordinates of P on each median.
Step 4:
Find the coordinates of P on the median that starts at vertex (0, 0) and ends at midpoint \((\frac{(a+b)}{2}, \frac{c}{2})\):
\(\frac{1}{3}(0, 0) + \frac{2}{3}(\frac{(a+b)}{2}, \frac{c}{2}) = (0 + \frac{(a+b)}{3}, 0 + \frac{c}{3}) = \left(\frac{(a+b)}{3}, \frac{c}{3}\right)\).
Find the coordinates of P on the median that starts at vertex (a, 0) and ends at midpoint \((\frac{b}{2}, \frac{c}{2})\):
\(\frac{1}{3}(a, 0) + \frac{2}{3}(\frac{b}{2}, \frac{c}{2}) = (\frac{a}{3} + \frac{b}{3}, \frac{0 + c}{3}) = \left(\frac{(a + b)}{3}, \frac{c}{3}\right)\).
Find the coordinates of P on the median that starts at vertex (b, c) and ends at midpoint \((\frac{a}{2}, 0)\):
\(\frac{1}{3}(b, c) + \frac{2}{3}(\frac{a}{2}, 0) = (\frac{b}{3} + \frac{a}{3}, \frac{c}{3}) = \left(\frac{(a + b)}{3}, \frac{c}{3}\right)\).
Step 5:
The coordinates of P on each median are \(\left(\frac{(a+b)}{3}, \frac{c}{3}\right)\), which proves that the three medians of this generic triangle all intersect at the same point.