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an arbitrary point P is chosen on side BC of triangle ABC and perpendiculars PU and PV are drawn from P to other two sides of the triangle. (It may be that U or V lies on an extension of AB or AC and not on the actual side of the triangle. This can happen, for instance, if angle A is obtuse and point P is very near B or C.) Show that the sum PU+PV of the lengths of the two perpendiculars is constant as P moves along BC. In other words, this quantity is independent of the choice of P.

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Please don't post problems in the imperative mode, as if you were giving assignments or homework to the group (clearly, you arecopying this from somewhere). If this is homework, please tag it with the [homework] tag. In any case, it is impolite not to phrase your question as a question; try to say why you are trying to solve this problem, what you've tried, what your thoughts are, and/or where you are stuck. –  Arturo Magidin Jan 30 '11 at 21:26

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Taking $P$ to be $B$ and $C$, we get that you actually need to assume that the triangle is isosceles ($AB = AC$). Given that, draw the segment $PA$ and express the area of the triangle in terms of $PU+PV$.

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I don't see how this is true: if you have a triangle such that $AB \neq AC$ then taking $ P = B$ and then $P = C$ gives you a contradiction. I think you at least need an isosceles triangle for this to work.

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