<s>[INST] 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
$PC^2=PE^2+z^2$
$PC^2=PF^2+(10-z^2)$
 [/INST] 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 $$PE+PD+PF=1+2+2=5$$
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