Construct $\triangle ABC$ in which $BC = 6 \; cm$, $\angle ABC = 120^\circ$ and $AB = 7.5 \; cm$
- Construct the locus of a point equidistant from $B$ and $C$.
- Draw the locus of a point $P$ such that area $\triangle PBC = $ area $\triangle ABC$.
$\triangle ABC$ is constructed with the given dimensions (red construction arcs).
Locus of a point equidistant from points $B$ and $C$ lies on the perpendicular bisector of $BC$ (line $DE$) (blue construction arcs and lines).
Locus of point $P$ [such that area $\triangle PBC = $ area $\triangle ABC$] lies on line $m$ which is a line through point $A$ and parallel to side $BC$ of $\triangle ABC$ (violet construction arcs and lines).