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)

Observe that, $\frac {d_{a}}{h_{a}}=\frac{\frac12\times a\times d_{a}}{\frac12\times a\times h_{a}}=\frac{Area(KBC)}{Area(ABC)}$.
Similarly, $\frac{d_{b}}{h_{b}}=\frac{Area(KCA)}{Area(ABC)}$ and $\frac{d_{c}}{h_{c}}=\frac{Area(KAB)}{Area(ABC)}$.
Adding these three ratios will give the desired result.
