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feet of perpendicular and midpoint

Posted: Mon Apr 04, 2016 2:02 pm
by RJHridi
In any triangle ABC, P, Q, R are the feet of perpendiculars from a point S to the sides BC, CA, AB respectively. L, M, N, are the midpoints of these sides. Prove that (LP)(BC) + (MQ)(CA) + (NR)(AB) = 0

Re: feet of perpendicular and midpoint

Posted: Wed Apr 27, 2016 4:17 pm
by asif e elahi
It will be equal to $0$ if you take directed segment, otherwise not.

Hint
Write $\vec{LP}.\vec{BC}$ in terms of $\vec{BP}$ and$\vec{CP}$.