Wednesday, May 18, 2016

Geometry Problem 1214: Triangle, Three Angle Bisectors, Proportions

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Click the figure below to view more details of problem 1214.

Geometry Problem 1214: Triangle, Three Angle Bisectors, Proportions

Posted by Antonio Gutierrez at 8:08 AM
Labels: , ,

4 comments:

  1. Jacob HA (EMK2000)May 18, 2016 at 9:00 AM

    c/a
    = AD/DC
    = (AD/BD) / (DC/BD)
    = ((c - e)/e) / ((a - f)/f)

    ace - cef = acf - aef
    1/f - 1/a = 1/e - 1/c
    1/a + 1/e = 1/c + 1/f

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    Replies
    1. AnonymousMay 24, 2016 at 7:46 AM

      c((a - f)/f) = a((c - e)/e)
      ((a - f)/f)/a = ((c - e)/e)/c
      1/f - 1/a = 1/e - 1/c
      1/a + 1/e = 1/c + 1/f

      Delete
  2. Sailendra TMay 18, 2016 at 9:05 AM

    Let AD = y; AC = x AND BD = z.
    Applying angle Bisector to triangle ABC => x/y = a/c --------- (1)
    Applying angle Bisector to triangle ABD => y/z = (c-e)/e ----------- (2)
    Applying angle Bisector to triangle BDC => x/z = (a-f)/f -----------(3)

    (3)/(2) => x/y = a/c = e.(a-f)/f.(c-e)
    => af(c-e) = ce(a-f)
    => a(c/e-1) = c(a/f-1)
    => 1/e-1/f = (a-c)/ac
    => 1/a+1?e = 1/c+1/f

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  3. c.t.e.oMay 18, 2016 at 12:30 PM

    1) c/AD=a/DC => 1.1)AD/DC=c/a
    2) (c-e)/AD=e/BD
    3)(a-f)/DC=f/BD
    from 2) & 3) 1.2) AD/DC=(c-e)f/(a-f)e
    from 1.1 &1.2 it's proved

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