How to find sum of 3 perpendiculars of a triangle? Q. ABC is an equilateral triangle with side 10cm and P is a point inside the triangle, at a distance of 2cm from AB. If PD, PE and PF are perpendiculars to the three sides, find sum PD+ PF+PE.


What I've done:
In triangles ADP and AFP,
$AP^2=x^2+PF^2$
$AP^2=y^2+4$
In DPB and PBE
$PB^2=4+(10-y)^2$
$PB^2=PE^2+(10-z)^2$
In PEC and PFC
$PC^2=PE^2+z^2$
$PC^2=PF^2+(10-z^2)$
 A: First let's introduce you to the plan. First we'll prove that the sum of the perpendiculars is constant and independent of the point $P$. Also we'll prove that their sum is equal to the height of the triangle.

(Note that the notation is little different, but you'll understand.)
Draw segments connecting P with each of the vertices. Now we have three triangles $\triangle APB, \triangle APC, \triangle CPB$. It easy to spot that none of these three triangles overlap each other so the the sum of their sum is equal to the sum of $\triangle ABC$, because they are stucked inside. Because $PD, PE$ and $PF$ are perpendicular to the side of $\triangle ABC$ they connect to, they are the heights of their respective triangles. Also let's denote the height of $\triangle ABC$ as $h$. So we have:
$$P_{APB} + P_{APC} + P_{BPC} =  P_{ABC}$$
Using the fact that all sides are equal and the height formula for area of triangle we have:
$$\frac{a \cdot PD}{2} + \frac{a \cdot PF}{2} + \frac{a \cdot PE}{2} = \frac{h \cdot a}{2}$$
Now divide by $\frac a2$ we get:
$$PD + PF + PE = h$$
So we proved that the sum of the perpendiculars to the side from each point is constant, because for a given triangle the height is constant. Actually this works for every point in an equilateral triangle, not just points that are $2cm$ from one side. Actually this is statement is called Viviani's Theorem.
The length of the line in an equilateral triangle is given by the formula:
$$h = \frac{a\sqrt{3}}{2} = \frac{10\sqrt{3}}{2} = 5\sqrt{3}$$
So we have: $$PD + PF + PE = 5\sqrt{3} \approx 8.6602540378443865$$
A: Viviani's Theorem takes care of this problem very quickly. It states that

Let $P$ be any interior point of an equilateral triangle. The total sum of the distances from $P$ to the 3 sides is equal to the altitude of the triangle.

So the answer to your question is simply the altitude of the triangle, which can be found to be equal to $5\sqrt{3}$ using basic trigonometry. 
So in other words, the result is true regardless of where $P$ was chosen, so long as it is inside the triangle.
A: From equilateral triangle and from $DB=FC$ we see $x=y$.
By using Carnot's theorem:
$$(10-z)^2+(10-y)^2+y^2=z^2+x^2+(10-y)^2$$
$$100-20z=x^2-y^2=0$$
$$5=z$$
From here we can say that the points $A,P,E$ are on a line and $AE=BE=EC=5$.
So $AP$ is bisector of $DAF$ and from there we can conclude that $FP=2$ and $AP=4$. 
Hence $$PE+PD+PF=1+2+2=5$$
