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Let G be the centroid of ABC. The midpoints of sides BC, CA and AB are L, M and N, respectively. Also, D is the foot of the altitude from A to BC and K is the foot of the altitude from L to MN. 

 

 

(a) Show that AD/LK = 2

 

(b) Show that triangle ADG ~ triangle LKG

 

(c) Show that D, G, and K are collinear and that DG/GK = 2

 Mar 26, 2020
 #1
avatar+632 
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a) Here

 

b) Angle AGD is congruent to Angle KGL as they are vertical angles. AD is parrelel to KL because they both are perpendicular to the same line. Because AD is parrelel to KL we know by Alternate Interior Angles that ADG is congruent to GLK. Since we know two angles are congruent, by AA similarity, the triangles are similar.

 

c) DG/GK = 2 because we already proved that AD/LK = 2. They are the sides of a similar triangles, so the sides are proportional.

 

Not sure how D, G, and K are collinear. CPhill help me!!! (I have to prove that the angles formed by DG and GK is 180 by I can't find a way!)

 Mar 26, 2020
 #2
avatar+1956 
+1

Awesome, dude!

CalTheGreat  Mar 26, 2020
 #3
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Thank! ►smiley

AnExtremelyLongName  Mar 26, 2020
 #4
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Thank you so much!!! ^^

Guest Mar 26, 2020
 #5
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Good job AELN   !!!!!

 

And.....welcome aboard   !!!!!!

 

 

cool cool cool

 Mar 26, 2020
 #6
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Um, can someone please help me with proving the collinear lines here? I'm stuck. Thanks

 Mar 26, 2020
 #7
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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!

 Apr 1, 2020

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