Solution to NRich problem, "Tri-split".

Labelling points as in the interactive diagram, and with a = PT, b = PU and c = PS.
Conjecture: a+b+c is constant as long as P remains within the triangle.
Consider the triangle PTC.  It is a right-angled triangle, with base TC.  Its height is a, since the side of length a is perpendicular to the base.
Therefore, its area is ½a*TC.
Similarly, the area of triangle PTB is ½a*TB; and similarly for the other four triangles PSA, PSB, PUA and PUC.
Hence, the total area of the six small triangles is
At = ½aTC + ½aTB + ½bUA + ½bUC + ½cSA + ½ cSB
Factorising,
At = ½(a(CT + TB) + b(AU + UC) + c(AS + SB))
But CT + TB = CB, and similarly for the others, giving
At = ½(aCB + bAC + cAB)
Now we apply two geometrical facts.  Firstly, since triangle ABC is equilateral, AB = AC = CB = some value which we will label 'k'.
Therefore At = ½(ak + bk + ck)
Secondly, the six small triangles collectively make up the large triangle ABC, and therefore At is also the area of ABC.
Since ABC does not change in area however much we move P, At is a constant.
Rearranging At = ½(ak + bk + ck) gives
2 * At / k = a + b + c
And since k and At are both constant, this means that a + b + c is a constant.
In fact, since At = ½ * height * base, and base is k, and height is k sin 60 which is k√3/2, At = k² √3 / 4.
Therefore a+b+c = 2 * (k² √3 / 4) / k = k√3 / 2.
In other words, the sum of the perpendicular distances from P to each of the sides of the triangle is constant, and equals the side length times one-half of the square root of 3.

This can be seen on the interactive diagram, where the side length is 200, and therefore the formula gives that a+b+c = 200√3 / 2 = 100√3 = 173.205 to 3 decimal places - which is the value given by the diagram.

Of course this result only holds as long as P is within the triangle, since otherwise the areas of some of the triangles would have to be subtracted rather than added.