+0  
 
0
884
3
avatar+879 

In right triangle ABC, Angle B=90 degrees, and  D and E  lie on AC such that \(\overline{BD}\) is a median and \(\overline{BE}\) is an altitude. If BD=2*DE, compute \(\frac{AB}{EC}.\).

 

https://latex.artofproblemsolving.com/3/4/8/3485794f7f5958afaee721bc44e35ed36950668e.png

 Mar 13, 2019
 #1
avatar+111437 
+2

 

Since  BD is a median....then AD = DC........and we can construct a circle with radius DC  = 2

And since angle B is right  and A and C lie on the diameter endpoints .....then right triangle ABC can be inscribed in the circle

 

Then BD is a radius   = 2.....and...in right triangle BDE,  DE = (1/2) BD = 1  ....and BE = √3

And in  right triangle BDE, angle DBE = 30°   and angle EDB = 60°

Then angle ADB is supplemental to EDB = 120°

And AD = DB.....so  in triangle ABD, angles BAD and ABD =  30°

Then, in right triangle ABC,  angle ACB = 60°

So....in  right triangle ABC....BC = (1/2) AC = (1/2)(4) = 2  and AB =  2√3

And in right triangle BCE....angle EBC = 30°....so....EC = 1/2 BC = (1/2)(2) = 1

 

So

 

AB / EC  =      [2√3] / [1]   =     =    2√3

 

 

cool cool cool

 Mar 13, 2019
 #2
avatar+110154 
0

Hi Chris,

I really like your idea of putting it in a semicircle!      cool

 Mar 13, 2019
 #3
avatar+111437 
+2

I learned that trick from Heureka......it's probably somewhat "gimmicky," but....any port in a storm....LOL!!!

 

 

cool cool cool

CPhill  Mar 13, 2019
edited by CPhill  Mar 13, 2019

21 Online Users

avatar
avatar
avatar
avatar