Your Friend’s Ratio “Mistake”

I'm not sure I understood the point here as the answer seems to be given in the question. Nevertheless....

We use the notation $A_{I}$ for the area of triangle $I$ , etc. As noted in the question, $A_{III} = A_{II} + A_{I}$ , because the triangle formed by combining triangles $II$ and $I$ has an equal base and the same height as triangle $III$ . Also as noted, triangles $I$ and $II$ have equal heights; but this means that the ratio of their areas will equal the ratio of their bases. So $\begin{aligned} A_{I} : A_{II} = 6 : 8 &\implies A_{I} : A_{II} + A_{I} = 6 : (8 + 6) = 3 : 7 \qquad \small \color{#3D99F6} \text{[and then, because } A_{III} = A_{II} + A_{I}] \\ &\implies A_{I} : A_{III} = 3 : 7 \end{aligned}$

After me and my friend ran into this problem we tried using the ratio properties of triangles we had learned to understand why this result happened. When two triangles share a height, the ratio of their areas is the same as the ratio of their bases. Therefore, we tried finding a way to prove that the altitude of $\triangle CED$ (in the diagram below) drawn to $\overline{AG}$ is congruent to the altitude of $\triangle ABD$ drawn to $\overline{AG}$ .

Finally before we make some real headway into the problem, we should formalize what we are even trying to prove, which is,

For any $\triangle ABC$ , let $\overline{AD}$ be the median drawn from point $A$ to $\overline{BC}$ . Let $E$ be any point on $\overline{AD}$ . The ratio of the areas $\triangle ABD$ and $\triangle CED$ is equal to the ratio of $\overline{AD}$ to $\overline{ED}$ .

Triangle Diagram

To prove this lets start by drawing altitudes $\overline{BH}$ and $\overline{CG}$ to their respective bases. To make it easier to visualize what's about to happen, lets move $\overline{CG}$ "up" along its base, keeping it perpendicular to $\overline{BG}$ , giving us $\overline{DF}$ (note that it's length remains the same as it forms a rectangle). Because $\overline{BH}$ and $\overline{DF}$ are both perpendicular to the same line ( $\overline{AG}$ ), $\overline{BH} \parallel \overline{DF}$ . Base $\overline{BC}$ is a transversal that cuts $\overline{BH}$ and $\overline{DF}$ , so $\angle HBD \cong \angle FDC$ , and because of the $AAS$ Congruency Theorem, $\triangle BHD \cong \triangle DFC$ . Then, the altitudes are congruent, so the ratio of areas is equal to the ratio of bases, and we've proved our conjecture.

It turns out that there is a reason for the relationship and the numbers didn't just happen to work out, pretty neat!