# Distances in a right triangle

• Mar 31st 2013, 11:30 PM
geniusgarvil
Distances in a right triangle
ABC is a right triangle with ABC=90 with AB=30 sqrt 3and BC=30. D is a point on segment BC such that AD is the median. E is a point on segment AC such that BE is perpendicular to AC. AD and BE intersect at F. If EF=(a sqrt b)/c, where aand c are positive, coprime integers and b is not divisible by the square of any prime, what is the value of a+b+c?
• Apr 1st 2013, 01:18 AM
MINOANMAN
Re: Distances in a right triangle
The triangle ABC is 30,90,60 triangle....with BE as height and AD as median...angle BAD is easy to calculate so does the angle FAE.. AC = 60 ,AE = 45 and EC = 15 .
You have whatever you need to find EF.....then compare it with what you have....just elementary algebra..
• Apr 17th 2013, 11:24 AM
johng
Re: Distances in a right triangle
Hi,
Everything in the previous posting is certainly true, but I don't see how this immediately gives the distance EF. Here is a purely algebraic solution:

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