Let's fill in the missing pieces for the proof step by step. Here’s how you would complete the proof:
Step 1: Draw the three medians of the triangle. Draw each median so that it starts at a vertex and ends at the midpoint of the opposite side.
Step 2: Find the coordinates of each midpoint by taking the sum of the x-coordinates and dividing by 2 and taking the sum of the y-coordinates and dividing by 2.
The midpoint between (0, 0) and (b, c) is \(\left(\frac{b}{2}, \frac{c}{2}\right)\). The midpoint of (0, 0) and (a, 0) is \(\left(\frac{a}{2}, 0\right)\). The midpoint of (a, 0) and (b, c) is \(\left(\frac{a+b}{2}, \frac{c}{2}\right)\).
Step 3: Find the coordinates of P, the point where the medians appear to intersect, on each median. Use the formula \(\frac{2}{3} \text{(midpoint)} + \frac{1}{3} \text{(vertex)}\).
Step 4: Find the coordinates of P on the median that starts at vertex (0, 0) and ends at midpoint \(\left(\frac{a+b}{2}, \frac{c}{2}\right)\).
\(\frac{1}{3}(0, 0) + \frac{2}{3}\left(\frac{a+b}{2}, \frac{c}{2}\right) = \left(\frac{0 + a + b}{3}, \frac{c}{3}\right) = \left(\frac{a+b}{3}, \frac{c}{3}\right)\)
Find the coordinates of P on the median that starts at vertex (a, 0) and ends at midpoint \(\left(\frac{b}{2}, \frac{c}{2}\right)\).
\(\frac{1}{3}(a, 0) + \frac{2}{3}\left(\frac{b}{2}, \frac{c}{2}\right) = \left(\frac{a}{3} + \frac{b}{3}, \frac{c}{3}\right) = \left(\frac{a+b}{3}, \frac{c}{3}\right)\)
Find the coordinates of P on the median that starts at vertex (b, c) and ends at midpoint \(\left(\frac{a}{2}, 0\right)\).
\(\frac{1}{3}(b, c) + \frac{2}{3}\left(\frac{a}{2}, 0\right) = \left(\frac{b}{3} + \frac{a}{3}, \frac{2c}{3}\right) = \left(\frac{a+b}{3}, \frac{c}{3}\right)\)
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.
This completes the proof.