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

 May 1, 2020
 #1
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(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

 

 

cool cool cool

 May 1, 2020
edited by CPhill  May 1, 2020
edited by CPhill  May 1, 2020
 #2
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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!

 May 3, 2020

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