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"Find the locus of a point P such that the sum of its distances from sides AB and AC of a Triangle ABC is equal to its distance from BC."
Is it possible to deduce the answer purely by geometry or are co-ordinates my only way out?
My earlier intuitions that the locus is a straight line is clear from co-ordinate geometry but I am not able to yet locate exactly where it lies (excluding the equation of a line in the x-y plane).
I also suggest the possibility of the locus being "the line joining the feet of angle bisectors from B and C onto AC and AB" as it holds true at the two points on AB and AC but I am not able to prove the same

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closed as off-topic by José Carlos Santos, Arnaud D., Guy Fsone, Raffaele, Parcly Taxel Dec 6 '17 at 3:12

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    $\begingroup$ The locus does have a simple geometric relationship to the triangle, but the proofs I've seen use coordinate geometry. Maybe you should try and see if you can figure out both the coordinate proof and the geometric significance. As a start, consider the points where the locus of $P$ intersects the sides $AB$ and $AC.$ $\endgroup$ – David K Dec 6 '17 at 2:36
  • $\begingroup$ My suggestion was that the locus turns out to be a straight line joining the feet of angle bisectors from B and C onto AC and AB but I am not able to prove the same $\endgroup$ – Dhvanit Dec 6 '17 at 13:28
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    $\begingroup$ It looks like you're really close to the solution. Remember that two points determine a line. One can show that the two points you found geometrically are on the line determined by the coordinate-based answer. This may also help: math.stackexchange.com/questions/2357255/… $\endgroup$ – David K Dec 6 '17 at 14:01
  • $\begingroup$ Thanks a lot! I am finally able to complete my geometric proof. $\endgroup$ – Dhvanit Dec 6 '17 at 16:09
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Let the vertices of triangle be ${\rm A}\left(x_{1},y_{1}\right),{\rm B}\left(x_{2},y_{2}\right),{\rm C}\left(x_{3},y_{3}\right)$ and the locus point be ${\rm R}\left(h,k\right)$

$${\rm AB}=\left(y-y_{1}\right)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)$$

So, the distance of $\rm R$ from line $\rm AB$ is

$$d_{1}=\frac{k-y_{1}-\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(h-x_{1}\right)}{\sqrt{1+\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)^{2}}}$$

Similarly, $d_{1}+d_{3}=d_{2}$ to get the locus of $\left(h,k\right)$.

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  • $\begingroup$ This might provide me with an answer that is a circle or a line (or maybe some random points) but by using coordinate geometry, the answer will have no significance with respect to the triangle. So, I would desire a purely geometric solution $\endgroup$ – Dhvanit Dec 5 '17 at 11:48

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