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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 $$\frac{DG}{GK}=2.$$

Thank you!

Feb 15, 2020

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(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!

Feb 15, 2020