Let \(G\) be the centroid of \(ABC.\) The midpoints of sides \(\overline{BC},\) \(\overline{CA},\)and \(\overline{AB},\) are \(L,\) \(M,\) and \(N\) respectively. Also, \(D\) is the foot of the altitude from \(A\) to \(\overline{BC},\) and \(K\) is the foot of the altitude from \(L\) to \(\overline{MN}.\)
Image -----> https://latex.artofproblemsolving.com/1/e/8/1e8da9ad587b4d838de98053bbb5844eb83decbf.png
(a) Show that \(\frac{AD}{LK}=2.\)
(b) Show that \(\triangle ADG\sim \triangle LKG.\)
(c) Show that \(D, G, and K\) are collinear and that \(\frac{DG}{GK}=2.\)
Thank you!
(a) Let P be the intersection of AD and MN, and let Q be the intersection of AL and MN. Then by Menelaus's Theorem, Q is the midpoint of PK. Since AD and LK are parallel, triangles ADL and LKQ are similar. And since Q is the midpoint of PK, QK = PK/2 = DL/2. Therefore, from the similar triangles, AD/LK = 2.
(b) Since QK = DL/2 and QG = LG/2, triangles GDL and GKQ are similar. Then KG = DG/2, so by part (a), triangles ADG and LKG are similar.
(c) In part (b), we found that triangle GDL and GKQ were similar. Then angle QKG = angle LDG, so D, G, and K are collinear. And by the similarity in part (b), DG/GK = 2. Easy!
