Wednesday, December 23, 2009

Problem 406. Right triangle, 15 degrees, Midpoints, Congruence

Proposed Problem
Click the figure below to see the complete problem 406 about Right triangle, 15 degrees, Midpoints, Congruence.

Problem 406. Right triangle, 15 degrees, Midpoints, Congruence.
See also:
Complete Problem 406
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 11:32 AM
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3 comments:

  1. Join G to B
    |GB|=1
    Join F to B
    |FB|=1
    see m(FBG)=60 and FBG is equilateral triangle
    |FG|=1

    ReplyDelete

  2. draw BM perpendicular to FG, GN perp to AC

    Right tr BMG and GNC are congruent ( BG=GC)=> GN = x/2
    but GN = 1/2 (ang 30) => x = 1

    ReplyDelete

  3. Connect BF and BG, AD=CE=2; AD and CE both hypotenuses. BF and BG are medians for them, so BF=BG=1. The median reveals Angle ABF=Angle CBG=15 Degrees. Angles (ABF+ABG+CBG)=90 Degrees, so Angle ABG=60 Degrees, which means FBG is equilateral.
    Therefore, FG=x=1.

    ReplyDelete

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