Lines $t$ and $s$ and point $P$ are given. Line $t$ is a perpendicular from the centroid of triangle $ABC$ to $BC$, line $s$ is a bisector of angle $\angle ABC$, and $P$ is the midpoint of $BC$. Construct triangle $ABC$.
I constructed a perpendicular from $P$ to $t$ and its intersection with the line s is point $B$. Point $C$ is the centrosymmetric image of point $B$ with respect to $P$, and point $A$ lies on the line $BP'$, where $P'$ is the axisymmetric image of point $P$ with respect to $s$. How do I find point $A$?
Best Answer
$\mathrm{Fig.\space 1}$ shows what you have done up to now. The given information is shown in $\color{red}{\text{red}}$. The black lines and points are what you have constructed. By the way, we renamed your axisymmetric image of point $P$ as $Q$.
$\mathrm{Fig.\space 2}$ shows what you should do to locate the vertex $A$ of $\triangle ABC$. Mark two arbitrary points $M$ and $N$ on the given line $t$ and draw the two rays $PM$ and $PN$. Now locate the point $K$ on $PM$ such that $MK=2PM$. Similarly, mark the point $L$ on $PN$ such that $LN=2PN$. To complete the construction join the points $K$ and $L$ to intersect the line $BQ$ as shown in the diagram. The point of intersection of $BQ$ and $KL$ is the sought vertex $A$.
I think that you are versed in geometry to figure out why the steps described above lead to what you were looking for.