ve that the perpendiculars to the sides at these points meet in common point if and only if $ BP^2 + CQ^2 + AR^2 = PC^2 + QA^2 + RB^2 $

Question

$P, Q, R$ are points on the sides $BC,CA,AB $ of triangle $ABC$. Prove that the perpendiculars to the sides at these points meet in common point if and only if

$ BP^2 + CQ^2 + AR^2 = PC^2 + QA^2 + RB^2 $

I can't seem to prove $ BP^2+CQ^2+AR^2 = PC^2+QA^2+RB^2 \Rightarrow $ "the perpendiculars to the sides at these points meet in common point".

I have proved the implication other way using Pythagoras theorem.

score 2 · Accepted Answer · answered Jun 30 '13 at 17:17

2

Let $X$ be the intersection of the perpendicular drawn from $Q$ and $R$. We have $$XB^2 - XC^2 = XB^2-XA^2 +XA^2-XC^2\\ =RB^2-RA^2+QA^2-QC^2\\ =PB^2-PC^2$$ From this you can show that $XP$ is perpendicular to $BC$.

answered Jun 30 '13 at 17:17

S.B.

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ve that the perpendiculars to the sides at these points meet in common point if and only if $ BP^2 + CQ^2 + AR^2 = PC^2 + QA^2 + RB^2 $

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