# 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.

• Thank you for the response! I really don't understand, though. The area is actually $S=\dfrac{ah_a}{2}$. Feb 1, 2021 at 17:16
• 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 Feb 1, 2021 at 17:18
• Oh, silly me! Thank you! 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.