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Equilateral $\triangle{ABC}$ with sides $2\sqrt{3}$. Let $P$ be the point outside$\triangle{ABC}$ such that points $A$ and $P$ lie opposite to $BC$. Let $PD$, $PE$, $PF$ be the perpendicular dropped on side $BC$, $AC$ and $AB$ receptively where $D$, foot of perpendicular, lies inside the segment $BC$. Let $PD=2$. How to find $PE+PF$.

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The sum of perpendiculars is altitude $H$ in case of an internal point P. If the point P lies outside and has length of nearest perpendicular $h$ then the total of perpendicular distances is:

$$ altitude \,H + h = 3 +2 =5, $$

which can be verified by drawing perpendiculars on all three sides and calculating extra length parts in $ PE,PF$,over and above that for point $P$ when $on$ nearest side $BC.$ Shown in red color in a rough sketch $PD=h$ below.

EDIT1:

Angles $$ DPE = DPF = 60^0 $$

Red length total = $$ h (\cos 60^0 + \cos 60^0) = 2 ( \frac12 + \frac12) =2 $$

Alt_EQ_tria

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  • $\begingroup$ how to calculate red part $=2$ ? $\endgroup$
    – mnulb
    Jul 17 '16 at 3:46
  • $\begingroup$ Please see above edit. $\endgroup$
    – Narasimham
    Jul 17 '16 at 4:57

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