1. **State the problem:** Given triangle $\triangle ABC$ with points $A(0,0)$, $B(0,a)$, and $C(a,0)$, prove that median $AD$ is also an altitude. 2. **Identify point $D$:** Since $D$ is the midpoint of $BC$, use the midpoint formula: $$D = \left(\frac{0+a}{2}, \frac{a+0}{2}\right) = \left(\frac{a}{2}, \frac{a}{2}\right)$$ 3. **Formula to use:** To prove $AD$ is an altitude, show that $AD$ is perpendicular to $BC$. Use the slope formula: $$\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}$$ 4. **Calculate slope of $BC$:** $$\text{slope}_{BC} = \frac{0 - a}{a - 0} = \frac{-a}{a} = -1$$ 5. **Calculate slope of $AD$:** Points $A(0,0)$ and $D\left(\frac{a}{2}, \frac{a}{2}\right)$: $$\text{slope}_{AD} = \frac{\frac{a}{2} - 0}{\frac{a}{2} - 0} = 1$$ 6. **Check perpendicularity:** Two lines are perpendicular if the product of their slopes is $-1$: $$\text{slope}_{AD} \times \text{slope}_{BC} = 1 \times (-1) = -1$$ 7. **Conclusion:** Since the slopes multiply to $-1$, $AD$ is perpendicular to $BC$, so median $AD$ is also an altitude. **Final answer:** Use the slope formula to show $AB \perp AC$ is incorrect here; the correct plan is to use the midpoint formula to find $D$ and then use the slope formula to show $AD \perp BC$. **Answer to multiple choice:** Use the midpoint formula to show point $D$ is the midpoint of $BC$.