Thursday, April 16, 2009

Problem 30. Right Triangle, Incenter, Inradius

Proposed Problem

Problem 30. Right Triangle, Incenter, Inradius.

See also: Complete Problem 30, Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 12:18 AM
Labels: , , ,

9 comments:

  1. AjitApril 22, 2009 at 9:15 PM

    From F draw FD perpendicular to AB meeting the latter in D. Now ED = r, angle EAB = angle EAF = A/2, r = AE sin(A/2) and AE = AFcos(A/2) which gives us: r/sin(A/2) = AFcos(A/2) or AF = r/sin(A/2)cos(A/2) = 2r/sin(A). Finally, BG = FD = AFsin(A) = 2r/sin(A) * sin(A) = 2r.
    Ajit: ajitathle@gmail.com

    ReplyDelete
  2. bae deok rakOctober 5, 2010 at 10:07 PM

    Let G be an intersecting point of the segment AB and the extension of FE and let
    P,Q and R be ponts on segments AB, BC and FG, respectively with EPㅗAB, EQㅗBC and ERㅗFG.
    Then we see that BQ=EP=EQ=r and
    two triangles EPG and ERF are congruence.
    Hence we get BG=BQ+QG=BQ+ER=BQ+EP=2r.

    ReplyDelete
  3. venkat sriramAugust 22, 2011 at 8:40 AM

    Let G be an intersecting point of the segment AB and the extension of FE and let
    P,Q and R be ponts on segments AB, BC and FG, respectively with EPㅗAB, EQㅗBC and ERㅗFG.
    Then we see that BQ=EP=EQ=r and
    two triangles EPG and ERF are congruence.
    Hence we get BG=BQ+QG=BQ+ER=BQ+EP=2r

    ReplyDelete
  4. PravinAugust 23, 2011 at 4:23 AM

    Through E draw MN ∥ AB with M on AF and N on BG.
    ∠AEM = ∠EAB = A/2 -> ME = MA
    ∠MEF = 90° - A/2 = ∠MFE -> ME = MF
    AB ∥ MEN ∥ FG
    So BN = NG
    But MEN ∥ AB -> EN ⊥ AB => ENB = EBN = 45°
    -> EN = BN
    So BN = NG = EN = r
    Hence BG = 2r
    (-> denotes "implies")

    ReplyDelete
  5. PravinAugust 23, 2011 at 10:16 PM

    Slight Correction:
    in the line
    "But MEN ∥ AB -> EN ⊥ AB => ENB = EBN = 45°"
    it should be read as
    " ∠BEN = ∠EBN = 45° "

    ReplyDelete
  6. AnonymousJanuary 16, 2015 at 2:07 PM

    Extend EF to cut AB at M; since M is the symmetrical of F about E and FG||AB, FG is tangent to the incircle and thus we are done.
    Best regards,
    Stan Fulger

    ReplyDelete
  7. Sumith PeirisAugust 11, 2015 at 9:56 AM

    Since FG is parallel to AB we can see that FE bisects < AFG. So E is the ex centre of Tr. FGC. Hence EG bisects right < BGF and so < EGB = 45. But < EBG = 45 and so BG= 2r where r is the in radius = altitude of isoceles Tr. BEG

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  8. Loukas SidiropoulosApril 25, 2016 at 11:31 AM

    https://www.facebook.com/photo.php?fbid=10206230384227198&set=a.10205987640598759.1073741831.1492805539&type=3&theater

    ReplyDelete
  9. AnonymousDecember 27, 2017 at 8:06 AM

    Very simple.
    AB and GF are both tangents to the incircle.
    Let X be a point such that AB⟂XE, let Y be a point such that GF⟂YE. Hence, XE and YE are both radii of the incircle. In fact, BGYX is a rectangle, since there are 4 right angles. Hence XY is a diameter.
    BG=XY and XY=2r, hence BG=2r
    proved.

    ReplyDelete

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