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In the acute triangle $ABC$ distances from internal point $S$ to sides $a,b,c$ are respectively $d_{a}, d_{b}, d_{c}$. Show that $\frac{d_{a}}{h_a}+\frac{d_b}{h_b}+\frac{d_c}{h_c}=1$ where $h_a, h_b, h_c$ are heights, such that $h_a$ starts on vertex $A$ and ends on side $a$, etc.

I thought to calculate it using areas, but I have problem with appliance it.

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  • $\begingroup$ Consider the triangles $ABS$, $BCS$, and $CAS$. $\endgroup$
    – Daniel Fischer
    Feb 9, 2014 at 14:33
  • $\begingroup$ thanks for hint, now I'am able to tackle this one $\endgroup$
    – Gregor
    Feb 9, 2014 at 14:59

1 Answer 1

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Hint: Let $A$ denote the area of the triangle. Then $2A=ad_a+bd_b+cd_c$. OTOH $2A=ah_a$ thus $a= 2A/h_a$, similar for $b$ and $ c$.

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