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# In right triangle ABC, Angle B=90 degrees, and D and E lie on AC such that is a median and is an altitude. If

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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
+99441
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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

Mar 13, 2019
#2
+100042
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Hi Chris,

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

Mar 13, 2019
#3
+99441
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I learned that trick from Heureka......it's probably somewhat "gimmicky," but....any port in a storm....LOL!!!

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