euclidean-geometrygeometrylocus
Let ABC a triangle, find the set $\Gamma$ of points M so that $Area(\triangle MBC)=Area(\triangle ABC)$ without using dot product.
I treat $Area(\triangle ABC)$ as constant $ \delta$ and define $h$ as the altitude from the vertice A. If $M_0$ is point such that $d(M_0, (BC))=h$ then $\Gamma$ is the line passing through $M_0$ and parallel to $(BC)$.
Here the problem is $h$ is still undefined.
Reflect $M$ with respect to the sides of $ABC$. You get an hexagon whose area is $2S$:
The hexagon can be decomposed as the union between $A'B'C'$ (whose side lengths are $\sqrt{3}MA,\sqrt{3}MB,\sqrt{3}MC$) and the isosceles triangles $CA'B',BC'A',AB'C'$. It follows that
$$ 2S = \frac{\sqrt{3}}{4}(AM^2+BM^2+CM^2)+ 3 S'$$
where $S'$ is the area of a triangle with side lengths $MA,MB,MC$. By Weitzenbock's inequality $AM^2+BM^2+CM^2 \geq 4\sqrt{3}S'$, hence $S'\leq \frac{S}{3}$ as wanted.
My proof would be
$$\frac{\triangle MBC}{\triangle ABC} = \frac{MB}{AB} = \frac{ND}{AD} = \frac{\triangle NCD}{\triangle ACD}.$$
Since $\triangle ABC = \triangle ACD$, we are done.
Best Answer
Let $l_1||BC$ and $A\in l_1$.
Also, le $l_2$ be the line symmetric to $l_1$ respect to line $BC$.
Now, easy to see that for all $M\in l_1\cup l_2$ we have $S_{\Delta ABC}=S_{\Delta MBC}.$
Let $M\notin l_1\cup l_2$.
We have two cases.
In this case $S_{\Delta MBC}<S_{\Delta ABC}.$
In this case $S_{\Delta MBC}>S_{\Delta ABC}.$
We checked all points of the plane, which says that $l_1\cup l_2$ is a needed locus.