+235 Let ABCDEF be a convex hexagon. Let A', B', C', D', E', F' be the centroids of triangles FAB, ABC, BCD, CDE, DEF, EFA, respectively. (a) Show that every pair of opposite sides in hexagon A'B'C'D'E'F' (namely A'B' and D'E', B'C' and E'F', and C'D' and F'A') are parallel and equal in length. (b) Show that triangles A'C'E' and B'D'F' have equal areas.
+981 (a)
Let \(M\) be the midpoint of \(\overline{BC}\), connect \(A\) to \(M\). \(\overline {AM}\) must pass through \(G\) since \(G\) is the centroid and \(\overline {AM}\) is a median of \(\triangle ABC\). In addition, connect \(D\) to \(M\), \(\overline {DM}\) must pass through \(H\) since \(H\) is the centroid and \(\overline {DM}\) is the median of \(\triangle BCD\).
\(\because \frac{MG}{MA}=\frac13=\frac{MH}{MD} \text{ and } \angle AMD = \angle AMD \therefore\) by SAS similarity, \(\triangle MGH \sim \triangle MAD\). Therefore\( \overline {GH} \parallel \overline {AD} \text{ and } AD=3GH.\) Using the same method to prove \(\triangle NKJ \sim \triangle NAD\), we can conclude \(\overline {KJ} \parallel \overline {AD} \text{ and } AD=3KJ. \because AD=3GH \text{ and } AD = 3KJ \therefore GH=KJ\), similarly \(\because \overline {KJ} \parallel \overline {AD} \text{ and }\overline {GH} \parallel \overline {AD} \therefore \overline {KJ} \parallel\overline {GH}\).
The same method is used to prove every pair of opposite sides in the hexagon are parallel and equal in length.
(b)
Let the parallel line of \(\overline{HI}\) through \(G\) and the parallel line of \(\overline{GH}\) through \(I\) intersect at . \(\because \overline{GH} \parallel \overline {OI} \text { and } \overline {GO} \parallel \overline{HI} \therefore GHIO\) is a parallelogram. Then, we can conclude \([GHI]=[GOI]\). \(\because \overline{GH} \parallel \overline {OI} \text{ and } \overline {KJ} \parallel\overline {GH} \therefore \overline {OI} \parallel \overline {KJ} \because GH = OI \text { and } GH = KJ \therefore OI = KJ\).
From this, we can conclude \(OIJK\) is also a parallelogram, which means \([IOK]=[IJK]\). \(\because \overline{HI} \parallel \overline {GO} \text{ and } \overline {HI} \parallel\overline {LK} \therefore \overline {GO} \parallel \overline {LK} \because HI = GO \text { and } HI=LK \therefore GO = LK\). From this, we can conclude \(GOKL\) is also a parallelogram, which means \([GOK]=[GLK]\). Therefore, \([GHIJKL]=2[GIK]\). Using the same method, we can conclude \([GHIJKL]=2[LHJ]\), which means that \([LHJ]=[GIK]\).
I hope this helped,
Gavin.
