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Question-16803




Question Number 16803 by ajfour last updated on 26/Jun/17
Commented by ajfour last updated on 26/Jun/17
Prove that in an equilateral   triangle the sum of the distances  from any point to the three sides  is a constant.
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{in}\:\mathrm{an}\:\mathrm{equilateral}\: \\ $$$$\mathrm{triangle}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{distances} \\ $$$$\mathrm{from}\:\mathrm{any}\:\mathrm{point}\:\mathrm{to}\:\mathrm{the}\:\mathrm{three}\:\mathrm{sides} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{constant}. \\ $$
Answered by mrW1 last updated on 26/Jun/17
Area of ΔABC=Δ  AB=BC=AC=a  Δ=(1/2)AB×PN+(1/2)AC×PM+(1/2)BC×PL  =(1/2)a(PL+PM+PN)  ⇒PL+PM+PN=((2Δ)/a)=constant
$$\mathrm{Area}\:\mathrm{of}\:\Delta\mathrm{ABC}=\Delta \\ $$$$\mathrm{AB}=\mathrm{BC}=\mathrm{AC}=\mathrm{a} \\ $$$$\Delta=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{AB}×\mathrm{PN}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{AC}×\mathrm{PM}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{BC}×\mathrm{PL} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{a}\left(\mathrm{PL}+\mathrm{PM}+\mathrm{PN}\right) \\ $$$$\Rightarrow\mathrm{PL}+\mathrm{PM}+\mathrm{PN}=\frac{\mathrm{2}\Delta}{\mathrm{a}}=\mathrm{constant} \\ $$
Commented by ajfour last updated on 26/Jun/17
Thank you Sir, didnot realise it  was as straight.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir},\:\mathrm{didnot}\:\mathrm{realise}\:\mathrm{it} \\ $$$$\mathrm{was}\:\mathrm{as}\:\mathrm{straight}. \\ $$

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