Let ABC be a triangle with centroid G. Points L, M, and N are the midpoints of sides BC, CA, and AB respectively. Let D be the foot of the altitude from A to BC and let K be the foot of the altitude from L to MN.
(a) Show that AD/LK=2
(b) Show that triangleADG is similar to triangleLKG
(c) Show that D, G and K are collinear and that DG/GK=2
The altitude and the median from vertex A of triangle ABC are 4 and 5 units long, respectively. The anlge determined by AB and the median is bisected by the altitude. Find AC.
(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!