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
Labels: , , ,

6 comments:

  1. AnonymousDecember 23, 2009 at 3:13 PM

    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. c .t . e. oDecember 24, 2009 at 2:39 PM

    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. AnonymousOctober 23, 2010 at 4:25 PM

    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
  4. Sumith PeirisSeptember 29, 2015 at 3:18 AM

    Solved in a very complex manner but Newzad's solution is so simple.

    ReplyDelete
  5. rv.littlemanMarch 10, 2022 at 9:43 AM

    See the drawing

    ∠BAD= π/2 => B, A and D are concyclic, circle with center F middle of BD
    => FB=FA=FD and ∠BAF=15°
    In the same way, GC=GA=GE and ∠CAG=15°
    =>∠FAG=90-15-15=60°
    ΔFAG is isosceles in A (AF=AG) and ∠FAG=60° => ΔFAG is equilateral
    Therefore x=FG=BD/2=CE/2

    ReplyDelete
  6. MarcoDecember 8, 2023 at 8:21 PM

    AD=CE (given)
    triangle BAD congruent to triangle triangle BCE (AAS)
    =>BA=BC & BD=BE
    AE=BA-BE
    CD=BC-BD
    =>AE=CD

    Let intersecting point of AD & EC be H
    AE=CE (proved)
    triangle EAH congruent to triangle DCH (AAS)
    =>AH=CH
    <HAC=<HCA
    <EAC=<DCA=45

    In triangle ADC
    <ADC=105
    sin105/AC=sin45/2
    AC=2sin105/sin45
    AC=2sqrt2*sin105
    AC=sqrt3+1

    In triangle AHC
    <AHC=120
    sin120/AC=sin30/AH
    AH=ACsin30/sin120
    AH=AC/sqrt3=(sqrt3+1)/sqrt3

    Triangle FHG ~ Triangle AHC
    x/AC=FH/AH
    x/AC=(AH-1)/AH
    x/AC=1/(sqrt3+1)
    x=1

    ReplyDelete

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