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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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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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