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 $PE+PF$.


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.


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

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


  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.