For a point K inside a triangle show an equality Let ABC be a triangle whose heights are $h_a,h_b$ and $h_c$. Let $K$ be any point inside the triangle, and $d_a,d_b$ and $d_c$ the distances of $K$ from the sides $a,b$ and $c$, respectively. Show that $$\dfrac{d_a}{h_a}+\dfrac{d_b}{h_b}+\dfrac{d_c}{h_c}=1.$$

We can draw a line from $K$ to each of $A, B,$ and $C$, forming three triangles $KAB, KBC,$ and $KCA$. We know $$S_{KBC}+S_{KAC}+S_{KAB}=S_{ABC}$$ or $$\dfrac{ad_a}{2}+\dfrac{bd_b}{2}+\dfrac{cd_c}{2}=\dfrac{ah_a}{2}=\dfrac{bh_b}{2}=\dfrac{ch_c}{2}.$$ If we multiply the last by 2, we get $$ad_a+bd_b+cd_c=ah_a=bh_b=ch_c.$$ I don't see anything else. Any help would be appreciated. (I would love to hear your thoughts on the problem)

Now from your last equation, call
$M=ah_a=bh_b=ch_c$.
Then you have that $\frac{ad_a+bd_b+cd_c}{M}=1$, that is, $\frac{ad_a}{M}+\frac{bd_b}{M}+\frac{cd_c}{M}=1$, that is, $\frac{ad_a}{ah_a}+\frac{bd_b}{bh_b}+\frac{cd_c}{ch_c}=1$, which is what you want.
