Sunday, November 23, 2008

Elearn Geometry Problem 210: Triangle, Angles, Auxiliary lines

Triangle, angles

See complete Problem 210 at:
gogeometry.com/problem/p210_triangle_angles_auxiliary_line.htm

Triangle, Angles, Auxiliary Lines. Level: High School, SAT Prep, College geometry

15 comments:

  1. the solve 30
    [img]http://www.al3ez.net/upload/d/ibrahim_ibrahim_page4.jpg[/img]

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  2. The angle between the auxiliary line BE and the segment BD is 30º. Does the data is enought o solve this problem ??

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  3. Okay Antonio. Thanks, I could to sove this problem. So, All data are correct !
    angle DBC + angle DEC = 180º. So, quadrilateral BCDE is inscribed. Now, the angle between the auxiliary line BE and the segment BD is 30º. Then, the angle DBE = DCE = a = 30°. On the other hand, angle EDC(=X)is equal to EBC = 90 -2a. So, X= 90 - 2(30)= 30º.

    Bolchoi(from Brazil)

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  4. Dear bolchoi,
    Thank you for your comments.
    My answer:
    1. Why angle DBC + angle DEC = 180º?
    2. You don't need more data to solve this problem.
    3. In fact, you can deduce (it's not a data) that the angle between BE and BD measured 30 degrees.

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  5. Dear Jandir,
    Thank you for your comments.
    My answer:
    1. Why angle DBC + angle DEC = 180º?
    2. You don't need more data to solve this problem.
    3. In fact, you can deduce (it's not a data) that the angle between BE and BD measured 30 degrees.

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  6. Dear Antonio,

    In fact I can`t to say at first that angle DBC + angle DEC = 180º. So, my answear at first should be wrong. I`l search better.

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  7. Firstly, I trace the line BE. Because the triangle BCE is isosceles(CB = CE), I can deduce that angle DBE = 30º. Now, Let M be a point on the midle point of segment BE(MB = ME)and let N be a point on BD. Then, CM is the height of tringle BCE. Now, because I can deduce that the angle MEN = 30º = DBE, I see that the triangle BNE is isoceles(NB = NE). Now I can deduce that the points B, C, E and N delimit the quadrilateral BCEN which is an inscribed quadrilateral inside an circle. As the points C, M and N are on the same line and the CM is the height of the isosceles triangle BCE, the line CN is the diameter of this circle. Then, the angle NBC = 90º and ABC is a right(or pytagorean) triangle. So, 120 - 2alpha = 90º and then alpha = 15º. Then, I can easily deduce that DBC is an right triangle(BC = BD and angle BDC = angle BCD =45º)and that BCE is an equilateral triangle. Now, I can see that triangle BDE is isosceles(BD = BE). As the angle DBE = 30º, then the angle EDB = 75º. Finally, the measure of the angle CDE = angle EDB - angle BDC = 75º - 45º = 30º

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  8. why #BCEN is an inscribed quadrilateral?

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  9. From where did you obtain BCEN is an inscribed quadrilateral ? , and why did you say "let N be a point on BD" , and afterwards you defined it with the angle MEN = 30º ? ok , let's say no problem there but , I repeat from W(here)TF did you obtain BCEN is an inscribed quadrilateral , it's an particular case , i think the real answer is : 2*ALPHA .
    Check other cases and it's not 30 degrees .

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  10. http://geometri-problemleri.blogspot.com/2009/10/problem-39-ve-cozumu.html

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  11. fie N pe BD astfel iancat CN=a.Se obtine triunghiul NEC echilateral, triunghiul DNC isoscel, triunghiul DNE isoscel si se calculeaza unghiul EDC=30.

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  12. Antonio,
    If alpha=15 deg. then clearly x=30 deg. and this can be done w/o trigonometry.
    However, in the general case, is x =2*(alpha)? And how may we prove it?
    Please enlighten us.

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    Replies
    1. Dear Ajit, Thanks for your comment. In general case x = 30 degrees. Hint for a geometric prove: draw CF = CB (F on AB extended)

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  13. Problem 210
    Let point P in the extension of AB to B such that CB=CP=CE (if α<15).Τhen 15.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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