Answer to Question #140802 in Geometry for Ali Atwi

Question

ABC and VDC are two isosceles triangles drawn on both sides of BC .E is the symmetric of C with respect to A and F is the symmetric of C with respect to D.
Show that E,B and F are collinear

Answer 1

Since $E$ is symmetric With respect to $A$ , $AC=AE$ , similarly to $CB=AB$ .

Let's lower the medians from the vertices $A$ and $D$ to the base $BC$ , these will be the medians $AH$ and $DH$ , so $AD$ is perpendicular to $CB$ and divides it in half. $AD$ is the middle line of $\triangle ECF$ (since $A$ and $D$ are the midpoints of $EC$ and $CF$ , respectively), hence $CB$ is perpendicular to $EF$, since the ratio of the triangles $\frac{\triangle CAD}{\triangle CEF} = \frac{1}{2}$ , and $\frac{CH}{CB}=\frac{1}{2}$ , then B belongs to the straight $EF$ , hence $B, E, F$ are collinear.