Results 1 to 8 of 8

Math Help - hard question. need geometric proof.

  1. #1
    Senior Member abhishekkgp's Avatar
    Joined
    Jan 2011
    From
    India
    Posts
    495
    Thanks
    1

    hard question. need geometric proof.

    ABC is an acute angled triangle. \angle BCA=80 $^{\circ}$. The angle bisector of \angle BCA meets AB in D. Let the perpendicular bisector of side AB meet the angle bisector of \angle BDC in E. Join E and C. Its given that CE bisects exterior \angle DCA (i.e. \angle DCE=70$^{\circ}$).
    Prove that \angle CBA=20$^{\circ}$

    The figure i have attached is not very clear but you do not need the figure. the text i have written is sufficient to understand the question.
    Attached Thumbnails Attached Thumbnails hard question. need geometric proof.-geomques.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jun 2010
    Posts
    205
    Thanks
    1
    I'm still working on this, but angle DCE is 90. Angle BCA is 80, so the exterior is 100, so each half is 50, and so on.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member BAdhi's Avatar
    Joined
    Oct 2010
    From
    Gampaha, Sri Lanka
    Posts
    252
    Thanks
    6
    ragnar,
    i think what abishek mentioning is that DC and the extending line is bisected by the line CE not BC and the extending line so i think conclusion- "each half is 50" is wrong,isn't it?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member BAdhi's Avatar
    Joined
    Oct 2010
    From
    Gampaha, Sri Lanka
    Posts
    252
    Thanks
    6
    my solution,

    if the \^{B}=x$^{\circ}$

    let the point intersecting EF and BC be M and the point intersecting ED and BC be N.

    since \^{F} = 90$^{\circ}$ \angle NMF = (90+x)$^{\circ}$

    consider point D, since \^{A}=(100-x)$^{\circ}$ \angle EDF=(70-\frac{x}{2})$^{\circ}$

    from \Delta CDN you'll find \angle MND=(80+x)$^{\circ}$

    now sum up all the angles in the quadrilateral and find x

    the answer will be 20
    Last edited by BAdhi; January 28th 2011 at 07:15 AM. Reason: previous post was repeated, so added new post
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jun 2010
    Posts
    205
    Thanks
    1
    Ah, right, good catch.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,561
    Thanks
    785
    Quote Originally Posted by BAdhi View Post
    from \Delta CDN you'll find \angle MND=(80+x)$^{\circ}$
    Could you explain this a bit more? I see that \angle MND is external to \triangle CDN, so \angle MND=\angle NCD+\angle CDN=40+\angle NDF=40+70-x/2=110-x/2. How do we get that it is 80+x? Also, did you use the fact that AF = FB?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member BAdhi's Avatar
    Joined
    Oct 2010
    From
    Gampaha, Sri Lanka
    Posts
    252
    Thanks
    6
    sorry, my mistake.

    took \angleNDC=40+x -totally idiotic
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member abhishekkgp's Avatar
    Joined
    Jan 2011
    From
    India
    Posts
    495
    Thanks
    1
    Quote Originally Posted by BAdhi View Post

    from \Delta CDN you'll find \angle MND=(80+x)$^{\circ}$
    hey badhi..
    how do you get this??
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: May 23rd 2011, 03:36 AM
  2. Sequence Proof (Hard)
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: October 7th 2009, 01:20 AM
  3. A hard limit proof
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: September 14th 2009, 01:22 AM
  4. Hard proof.
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: September 3rd 2009, 07:26 PM
  5. Hard Proof
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: March 8th 2008, 09:32 AM

Search Tags


/mathhelpforum @mathhelpforum