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 $\overline{MN}.$

(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 $\frac{DG}{GK}=2.$

Hi121529 May 1, 2020

#1**+1 **

(a) Show that $\frac{AD}{LK}=2

Since N and M divide sides BA and CA in the same proportion.....then NM is parallel to BC

Thus, triangle NAM is similar to triangle BAC

And since AN = 1/2 of AB and similar triangles are similar in all parts, then the altitude of triangle NAM =1/2 altitude of triangle BAC

But altitude of triangle NAM = altitude of triangle NML = KL

And the altitude of triangle BAC = AD

So

altitude of BAC / altitude of NAM = 2/1

Which implies that

AD / KL = 2 /1 = 2

CPhill May 1, 2020

#2**0 **

(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!

Guest May 3, 2020