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
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