Question 419848
{{{drawing(400,400,0,10,0,10,


triangle(2,2,8,2,5,2+3sqrt(3)),
locate(2,2,A),
locate(8,2,B),
locate(5,2+3sqrt(3), C),
locate(4, 3, D),
line(4,3,2,2),
line(4,3,8,2),
line(4,3,5,2+3sqrt(3))
)
}}}

It's a little hard to draw here, but let x,y,z be the altitudes from D onto AB, BC, CA, respectively (segment x is perpendicular to AB, etc.). We know that the sum of the areas of the triangles ADB, BDC, CDA add up to the area of ABC, so


{{{(x*AB)/2 + (y*BC)/2 = (z*CA)/2 = (h*AB)/2}}} where h is the altitude from C onto AB.


Since the triangle is equilateral, {{{AB = BC = CA}}} and we can replace BC, CA with AB, without loss of generality.


{{{(x*AB)/2 + (y*AB)/2 + (z*AB)/2 = (h*AB)/2}}}


We can multiply both sides by {{{2/AB}}}, cancelling out the AB and 2. Hence, we obtain


{{{x+y+z = h}}}, as desired.