Is HHH a congurence criteria for triangles?

Question

I wanted to know if a triangle defined by its 3 heights is unique. I took this up as a challenge but was able to get nowhere, can anyone help me? :)

Answer 1

Yes. Use SSS to construct the triangle with sides $\frac1{h_a}$, $\frac1{h_b}$, $\frac1{h_c}$. Then in this triangle the heights are proportinal to the given heights. Scale accordingly.

Answer 2

It is.

We are given any triangle with heights $h_a, h_b, h_c$, and we assume its sides to be $a,b,c$. To prove it is the only such triangle, we note that by the area formulae: $$a=\frac{2\triangle}{h_a}, b=\frac{2\triangle}{h_b}, c=\frac{2\triangle}{h_c}$$

We now use the fact that if a triangle with sides $a,b,c$ has area $K$ then area of the triangle with sides $ma,mb,mc$ has the area $Km^2$ (Do you see why?).

We introduce the notation that $[a,b,c]$ means the area of the triangle with sides $a,b,c$. So the previous fact could be written as $[ma,mb,mc]=m^2[a,b,c]$ Thus, the triangle with sides $a,b,c$ has the area $\triangle$ then:

$$ \triangle=[a,b,c] \\ = [\frac{2\triangle}{h_a}, \frac{2\triangle}{h_a}, \frac{2\triangle}{h_a}] \\ =4\triangle ^2 [\frac1{h_a},\frac1{h_b},\frac1{h_c}] $$

Thus $\triangle=\triangle^2 \times4[\frac1{h_a},\frac1{h_b},\frac1{h_c}]$, or : $$\triangle =\frac1{4[\frac1{h_a},\frac1{h_b},\frac1{h_c}]}$$

Thus, we have proved that the area can be deduced only from the heights independent of the sides. But now that we have the area and the heights, the sides are trivial to get by the area formulae. This proves that the triangle is unique, by the SSS criteria.

Additionally, we now have a method to get the sides of the triangle, and the angles, ... follow!