Prove base times altitude is a constant without resorting to area.

Question

The area of a triangle is one-half base times altitude. This implies that, for $\bigtriangleup ABC$, $ah_A = bh_B=ch_C$, where $h_A$ is the length of the altitude dropped from point A to side BC, etc.

I want to verify that this can be proved without using the concept of area.

So far all I've managed to do is verify Heron's formula, which can be interpreted as a proof that $ah_A = bh_B=ch_C = 2\sqrt{s(s-a)(s-b)(s-c)}$. So now I know that it can be done. My complaint is that I only wanted to show that $ah_A = bh_B=ch_C$, not find the actual value. Does anyone have or know where I can find such a proof?

Answer 1

You can use similar triangles concept. Let AA₁, BB₁, CC₁ - altitudes, then:

1) triangels AA₁C is similar to BB₁C (two angles), so AA₁/BB₁ = AC/BC, or ah_a=bh_b

2) triangels BB₁A is similar to CC₁A (two angles), so BB₁/CC₁ = AC/AB, or bh_b=ch_c

So ah_a=bh_b=ch_c