geometry
Inside an equilateral triangle $ABC$,an arbitrary point $P$ is taken from which the perpendiculars $PD,PE$ and $PF$ are dropped onto the sides $BC,CA$ and $AB$,respectively.Show that the ratio $\dfrac{PD+PE+PF}{BD+CE+AF}$ does not depend upon the choice of the point $P$ and find its value.
I have no idea how to proceed. I could not do anything beyond calculating the area of $PDB,PCE,PFA$ but here I get nothing. Please help!
Such a locus is a line segment: if $Q_A,Q_B,Q_C$ are the projections of a point $Q$ (inside $ABC$) on the sides $BC,AC,AB$ of the triangle, then: $$ QQ_A\cdot BC + QQ_B\cdot AC+QQ_C\cdot AB = 2[ABC]$$ holds, no matter where $Q$ is placed inside $ABC$. So, just find another point $P'$ on the perimeter of $ABC$ such that the sum of the distances from the sides $AB,BC$ equals $S$, and connect it with $P$: this will be your wanted locus. If you take signed distances, then the locus is the whole $PP'$-line.
If you like to use coordinate geometry, just put a skew reference system with centre in $B$ and the $x$-axis and $y$-axis oriented like $BA$ and $BC$. You are just solving $x+y=S$, that is the equation of a line.
First, we simplify the problem by allowing for signed distances, and consider points outside the triangle.
We will show that the set of points that satisfy $ax+by+cz=k$ is either the null set, a line or the entire plane. Then, when restricted to the (interior) of the triangle, this becomes a null set, a line segment, or the entire interior, so the result follows.
Hint: For a point $P$, if the side lengths of the triangle are $l_a, l_b, l_c$, then $l_a x + l_b y + l_c z = 2 \Delta $.
Hint: Hence, it follows that $ax+by+cz=k$ is the entire plane iff $a:b:c:k = l_a : l_b: l_c : 2 \Delta$.
If $a:b:c = l_a : l_b: l_c $, then we either get the entire plane or the null set.
Henceforth, assume $a:b:c \neq l_a : l_b: l_c $.
Hint: For fixed values of $a, b, c, k$, if $P_1$ and $P_2$ are distinct points that satisfy the equation, then so does the entire line segment $P_1 P_2$.
Corollary: The solution set is a (affine) sub-space. It remains to show that the solution set cannot be a single point.
Hint: For fixed values of $a, b, c, k$, if there is (at least) 1 point that satisfies it, then there are at least 2 distinct points that satisfy it.
Suppose we have the solution $P_1 = (x_1, y_1, z_1)$, then the point $P_2 = (x_1 + bl_c - cl_b, y_1 +cl_a - al_c, z_1 + al_b-bl_a)$ will also satisfy both equations (expand and cancel terms). These are distinct points if $a:b:c \neq l_a : l_b: l_c $.
Best Answer
For simplicity, let the sides of the triangle be $1$.
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Looking at the areas of the sub-triangles, we get $$ \begin{align} |\,\triangle ABP\,|&=\frac12|\,\overline{FP}\,|\times|\,\overline{AB}\,|=\frac12|\,\overline{FP}\,|\\ |\,\triangle BCP\,|&=\frac12|\,\overline{DP}\,|\times|\,\overline{BC}\,|=\frac12|\,\overline{DP}\,|\tag{1}\\ |\,\triangle CAP\,|&=\frac12|\,\overline{EP}\,|\times|\,\overline{CA}\,|=\frac12|\,\overline{EP}\,| \end{align} $$ Summing these, we get $$ |\,\overline{FP}\,|+|\,\overline{DP}\,|+|\,\overline{EP}\,|=2|\,\triangle ABC\,|=\frac{\sqrt3}{2}\tag{2} $$
Rename the distances in question as $x$, $y$, and $z$.
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Considering the vertical distances on sides $x$ and $y$ and then repeating the same for the other side pairs: $$ \begin{align} \frac{\sqrt3}{2}x+\frac12h_x&=\frac{\sqrt3}{2}(1-y)+\frac12h_y\\ \frac{\sqrt3}{2}y+\frac12h_y&=\frac{\sqrt3}{2}(1-z)+\frac12h_z\tag{3}\\ \frac{\sqrt3}{2}z+\frac12h_z&=\frac{\sqrt3}{2}(1-x)+\frac12h_x \end{align} $$ Adding these and cancelling results in $$ x+y+z=\frac32\tag{4} $$ That is $$ |\,\overline{FA}\,|+|\,\overline{DB}\,|+|\,\overline{EC}\,|=\frac32\tag{5} $$
Thus, $(2)$ and $(5)$ yield $$ \frac{|\,\overline{FP}\,|+|\,\overline{DP}\,|+|\,\overline{EP}\,|}{|\,\overline{FA}\,|+|\,\overline{DB}\,|+|\,\overline{EC}\,|}=\frac1{\sqrt3}\tag{6} $$