When two triangles have the same heights?

Question

Let $\Delta ABC$ and $\Delta EFD$ be triangles with sides $a$, $b$, $c$ and $d$, $e$, $f$. All sides have different lengths.

I am looking for a functional condition for sides when triangles can have the same heights $h_E=h_B$.

My attempt.

I have found the three cases when two triangles can have the same hights.

Case 1. Acute triangles

Case 2. Acute triangle and obtuse triangle

Case 3. Obtuse triangles

Answer 1

Since the area of the triangle is half the base length times the height, you can compute the area and divide by the base length to compare the heights.

If you only want to use the edge lengths, use Heron's formula to get $$A_1 = \frac{1}{4} \sqrt{(a + b + c)(-a + b + c)(a - b + c)(a + b - c)}$$ $$A_2 = \frac{1}{4}\sqrt{(d + e + f)(-d + e + f)(d - e + f)(d + e - f)}$$

So two triangles have the same height if $$\frac{\sqrt{(a + b + c)(-a + b + c)(a - b + c)(a + b - c)}}{b} = \frac{\sqrt{(d + e + f)(-d + e + f)(d - e + f)(d + e - f)}}{e}$$

Answer 2

Minimum distance between two parallel lines is the same, is the altitude $h$ of all triangles no matter where you choose the base and shift the base segment length AC or DF along this line AF and no matter where the apex B or E lies.