# acute triangle and internal point

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.

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

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 $c$.