Wednesday, August 25, 2010

Problem 511: Triangle, Double Angle, Angle Bisector, Measure

Geometry Problem
Click the figure below to see the complete problem 511 about Triangle, Double Angle, Angle Bisector, Measure.

Problem 511: Triangle, Double Angle, Angle Bisector, Measure, Mind Map.


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Complete Problem 511

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 9:27 AM
Labels: , , ,

9 comments:

  1. c .t . e. oAugust 25, 2010 at 11:41 AM

    three bisectors meet on G, BE = c (E on BC)
    => AGED rhombus => DEC isoceles => EC = ED = d
    => BC = c + d

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  2. Peter TranAugust 25, 2010 at 1:13 PM

    Draw CE perpendicular to BD and cut AB at F. ED cut BC at F.
    Since BD is angle bisector of angle (ABC) so triangle EBC is isosceles and BD become symmetrical line of triangle EBC.
    We have angle (BFD)=angle(BCA) =alpha (property of symmetric)
    And triangle EAD is isosceles. ( angle AED=angle ADE= alpha)
    AD=AE=BE-BA=BC-BA
    Peter Tran

    ReplyDelete
  3. geometriAugust 29, 2010 at 12:45 PM

    http://yfrog.com/mwp511doubleangletriangleg

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  4. geometriAugust 29, 2010 at 2:20 PM

    [URL=http://img822.imageshack.us/i/p511doubleangletriangle.gif/][IMG]http://img822.imageshack.us/img822/2781/p511doubleangletriangle.th.gif[/IMG][/URL]

    Uploaded with [URL=http://imageshack.us]ImageShack.us[/URL]

    ReplyDelete
  5. AjitAugust 29, 2010 at 5:54 PM

    Another construction: Extend BA to E s.t. AE=AD or /_AED=/_ADE=α. Triangles EBD & CBD are ASA congruent or BE=BC or c+d=a
    Ajit

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  6. INDSHASeptember 7, 2010 at 11:02 AM

    INDSHAMAT from University of Kelaniya,Srilanka..
    Draw the angle bisector of AngleBAC which cuts BC at E.Then AE=EC(as angleEAC=angleECA)
    d/DC=c/a(BD is the Bisector)So DC=ad/c (1)
    angleAEB=2@ and because angleABC is common for both ABE & ABC triangles,ABC & ABE triangles are similar.Hence AE/(DC+d)=CE/(DC+d)=c/a=BE/c (2).So BE=c2/a,hence CE=a-BE=(a2-c2)/a (3).From(2)we have CE/(DC+d)=c/a.So applying values of(3)&(1)to CE&DC we have a2-c2=d(a+c).So (a-c)(a+c)=d(a+c).So d=a-c

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  7. AnonymousApril 3, 2012 at 5:24 PM

    hello , this is another way to find the result :
    A' in BC such that : BA = BA' . from here we have : BA'D = 2x and AD = A'D .
    and we have to triangle A'DC is isosceles (DA'C = 2x and C=x) .
    hence we are done because : A'C = A'D = AD .
    adil azrou ...

    ReplyDelete
  8. Sumith PeirisJuly 13, 2015 at 7:29 AM

    Extend CA to E such that Tr. ABE is isoceles. Then Tr. EBC is isoceles from which it follows that Tr. BED is isoceles.
    Hence a = d+c so d = a - c.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  9. APOSTOLIS MANOLOUDISJune 21, 2016 at 5:39 AM

    Problem 511
    On the side BC I take the point E such that AB=BE, then triangle ABD=triangle EBD so AD=DE
    and <DEB=<DAB=2a. But <BED=<EDC+<ECD, so <EDC=a. Then EC=ED=AD=d.Therefore d=
    EC=BC-BE=a-c.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete

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