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 $$\dfrac{d_a}{h_a}+\dfrac{d_b}{h_b}+\dfrac{d_c}{h_c}=1.$$ enter image description here

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)


2 Answers 2


Now from your last equation, call


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.

  • $\begingroup$ Thank you for the response! I really don't understand, though. The area is actually $S=\dfrac{ah_a}{2}$. $\endgroup$
    – Kaloyan K.
    Feb 1, 2021 at 17:16
  • 1
    $\begingroup$ I called $S$ two times the area of the triangle. If you are more comfortable with it, I´ll call it by another name, $M$ for example $\endgroup$
    – Saúl RM
    Feb 1, 2021 at 17:18
  • $\begingroup$ Oh, silly me! Thank you! $\endgroup$
    – Kaloyan K.
    Feb 1, 2021 at 17:20

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.


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