Saturday, February 21, 2009

Problem 17: Right triangles and Angles

Proposed Problem
Problem 17: Right triangles and Angles.

See complete Problem 17 at:
gogeometry.com/problem/problem017.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

11 comments:

  1. From Tr. ABC, AB/AC = sin(2θ)
    AB = AC sin(2θ)=2ACsin(θ)cos(θ)
    = 2AC * DE/DC * DC/AC from Tr. CED & ADC
    Thus, AB = 2DE
    Ajit: ajitathle@gmail.com

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  2. This comment has been removed by a blog administrator.

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  3. http://ahmetelmas.files.wordpress.com/2010/05/cozumlu-ornekler.pdf

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  4. Nilton Lapa
    Take F on de extension of BC, so that CF=AC. Tr. ACF is isosceles, with angle(CAF)=angle(CFA)=(theta). Let BC be the median of ACF. Draw GH perpendicular to CF. DAC and GCF are congruent so GH=DE. As F is middle point of CF, we have GH=AB/2 and AB=2.DE.

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  5. AB/DE = AB/AC. AC/CD . CD/DE
    = sin 2θ. sec θ. cosec θ = 2

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  6. Where I said "Let BC be the median of ACF", please change to "Let CG be ..."
    Where I said "As F is middle point of CF", please change to "As G is middle poit of AF".
    Sorry about that.
    Nilton Lapa

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  7. Solution is uploaded to the following link

    https://docs.google.com/open?id=0B6XXCq92fLJJU2s3bDJBRDJBX2c

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  8. Take F - reflection of A in CD abd G projection of F onto AC. Easily FG=2DE and triangles CFG and CAB are congruent, hence FG=AB, done.

    Best regards,
    Stan Fulger

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  9. Let AC, DB meet at F and let AG drawn parallel to BC meet BD at G.

    ABCD is con cyclic and Tr.s ADF, BCF & AFG are all isoceles.

    Now Tr.s DEF & BAG being similar AB/DE = AG/EF = 2. AG/AF = 2 since AG=AF

    Hence AB = 2.DE

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  10. https://www.facebook.com/photo.php?fbid=10206193945516253&set=a.10205987640598759.1073741831.1492805539&type=3&theater

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  11. Another idea: take F reflection of D in E; AFBCD is cyclic and we need DF=AB. Since CF is angle bisector of <ACB, there exists the equality BF=AF; but as constructed AF=AD, thus DAFB is an isosceles trapezoid, having equal diagonals, DF=AB, done.

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