+0  
 
0
108
1
avatar

Rectangle ABCD has points E and F on sides AB and CD, respectively. If AE = AB/3 and CF = CD/4 and segments DE and BF intersect diagonal AC at G and H, respectively, what is the ratio AG:GH:HC? Express your answer in the form a:b:c, where a, b and c are relatively prime positive integers.

 May 8, 2021
 #1
avatar+121055 
+1

 

 

Triangles  AGE  and  CGD are similar

 

Since  AE = (1/3)AB

Then AE  = (1/3) CD

So AG  = (1/3) CG

CG = GH  + HC

 

And triangles  CHF and  AHB  are  similar

Then  CF  = (1/4)CD  = ( 1/4) AB

So  HC  = (1/4) HA

HC =  (1/4) ( GH  + AG)

 

AG  = (1/3)CG

AG =  (1/3) (GH  + HC)

3AG = GH + HC

HC  =  3AG - GH

 

Therefore

 

    HC  =  HC    so

3AG  - GH  =  (1/4)(GH +AG)

3AG - GH  =  (1/4)GH  + (1/4)AG

(3 - 1/4)AG  =  ( 1 + 1/4) GH

(11/4)AG  = (5/4)GH

11AG  = 5GH

AG  = (5/11)GH

 

And HC =  (1/4)(GH + AG)  =  (1/4)  [GH  + (5/11) GH ]   =  (1/4)(16/11)GH  =  (16/44)GH = (4/11)GH

 

So

 

AG   : GH  :  HC    =

 

(5/11)GH : GH : (4/11) GH   =

 

5GH   :  11 GH  :  4GH   = 

 

5  :   11   :   4

 

 

cool cool cool

 May 8, 2021
edited by CPhill  May 8, 2021
edited by CPhill  May 8, 2021

26 Online Users