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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

 Dec 8, 2019
 #1
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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.

 Dec 8, 2019
 #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!

 Dec 15, 2019

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