Sure! Here’s the completion of the proof using the provided steps.
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 \((\frac{b}{2}, \frac{c}{2})\). The midpoint of (0, 0) and (a, 0) is \((\frac{a}{2}, 0)\). The midpoint of (a, 0) and (b, c) is \((\frac{a+b}{2}, \frac{c}{2})\).
Step 3: Find the coordinates of P, the point where the medians appear to intersect, on each median. Use the formula
\(\frac{1}{3}\) vertex + \(\frac{2}{3}\) midpoint
Step 4: Find the coordinates of P on the median that starts at vertex (0, 0) and ends at midpoint \((\frac{a+b}{2}, \frac{c}{2})\).
\(\frac{1}{3}(0, 0) + \frac{2}{3}(\frac{a+b}{2}, \frac{c}{2}) = (0 + \frac{a+b}{3}, 0 + \frac{c}{3}) = (\frac{a+b}{3}, \frac{c}{3})\)
Find the coordinates of P on the median that starts at vertex (a, 0) and ends at midpoint \((\frac{b}{2}, \frac{c}{2})\).
\(\frac{1}{3}(a, 0) + \frac{2}{3}(\frac{b}{2}, \frac{c}{2}) = (\frac{a}{3} + \frac{b}{3}, 0 + \frac{c}{3}) = (\frac{a+b}{3}, \frac{c}{3})\)
Find the coordinates of P on the median that starts at vertex (b, c) and ends at midpoint \((\frac{a}{2}, 0)\).
\(\frac{1}{3}(b, c) + \frac{2}{3}(\frac{a}{2}, 0) = (\frac{b}{3} + \frac{a}{3}, \frac{c}{3}) = (\frac{a+b}{3}, \frac{c}{3})\)
Step 5: The coordinates of P on each median are \((\frac{a+b}{3}, \frac{c}{3})\), which proves that the three medians of this generic triangle all intersect at the same point.
Let me know if you need anything else!