Sure! Below are the missing pieces inserted into each step to complete the proof that all three medians of the triangle intersect at point \( P \).
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 \( \left(\frac{2x_1 + x_2}{3}, \frac{2y_1 + y_2}{3}\right) \).
Step 4: Find the coordinates of \( P \) on the median that starts at vertex (0, 0) and ends at midpoint \((\frac{a+b}{2}, \frac{c}{2})\). \[ \frac{1}{3}(0, 0) + \frac{2}{3}\left(\frac{a+b}{2}, \frac{c}{2}\right) = (0, 0) + \left(\frac{a+b}{3}, \frac{c}{3}\right) = \left(\frac{a+b}{3}, \frac{c}{3}\right) \]
Find the coordinates of \( P \) on the median that starts at vertex (a, 0) and ends at midpoint \((\frac{b}{2}, \frac{c}{2})\). \[ \frac{1}{3}(a, 0) + \frac{2}{3}\left(\frac{b}{2}, \frac{c}{2}\right) = \left(\frac{a}{3}, 0\right) + \left(\frac{b}{3}, \frac{c}{3}\right) = \left(\frac{a+b}{3}, \frac{c}{3}\right) \]
Find the coordinates of \( P \) on the median that starts at vertex (b, c) and ends at midpoint \((\frac{a}{2}, 0)\). \[ \frac{1}{3}(b, c) + \frac{2}{3}\left(\frac{a}{2}, 0\right) = \left(\frac{b}{3}, \frac{c}{3}\right) + \left(\frac{a}{3}, 0\right) = \left(\frac{a+b}{3}, \frac{c}{3}\right) \]
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.