AD,BE,CF are concurrent lines in a triangle ABC. Show that the lines through the midpoints of BC,CA,AB respectively parallel to AD,BE,CF are concurrent.
I am unable to proceed. Kindly comment on the technique to be used and state the geometrical theorems used.
Thanks in advance!
Source: Challenge and thrills of pre college mathematics.
Draw a picture. Let $M$, $N$, and $P$ be the midpoints of $BC$, $CA$, and $AB$ respectively. Then $\triangle MNP$ is similar to $\triangle ABC$ with parallel pair of edges. You can compare the angles $\angle BAD$, $\angle DAC$ with the corresponding angles at $M$. Similarly for the other angles at $B$ and $C$.
$ABC$ and its medial triangle are similar triangles: in particular, they have the same centroid $G$, and the medial triangle is the image of $ABC$ with respect to a dilation with centre $G$ and ratio $-\frac{1}{2}$.
Since homothetic transformations preserve parallelism, by assuming that $AD,BE,CF$ concur in $P$ we have that the parallel lines to $AD,BE,CF$ through the midpoints of $BC,AC,AB$ concur in $Q$, where $Q$ is the image of $P$ under a dilation with centre $G$ and ratio $-\frac{1}{2}$.
