How to find sum of 3 perpendiculars of a triangle?

Question

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 $PC^2=PE^2+z^2$

$PC^2=PF^2+(10-z^2)$

Answer 1

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 $$PE+PD+PF=1+2+2=5$$

Answer 2

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: $$PD + PF + PE = 5\sqrt{3} \approx 8.6602540378443865$$