Tuesday, December 8, 2009

Problem 400. Triangle, Angle bisector, Circumcircle, Perpendicular, Congruence

Proposed Problem
Click the figure below to see the complete problem 400 about Triangle, Angle bisector, Circumcircle, Perpendicular, Congruence.

Problem 400. Triangle, Angle bisector, Circumcircle, Perpendicular, Congruence.
See also:
Complete Problem 400
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 8:42 PM
Labels: , , , ,

11 comments:

  1. AnonymousDecember 9, 2009 at 1:41 AM

    |OG|=|OH|
    |OC|=|OD|=r
    |GC|=|DH|

    ReplyDelete
  2. Antonio GutierrezDecember 9, 2009 at 6:13 AM

    Reasons for step 1: |OG|=|OH| ?

    ReplyDelete
  3. c .t . e. oDecember 9, 2009 at 6:39 AM

    suggest to others

    1) draw HP perpendicular to AC ( P on AC )
    2) draw OK perpendicular to HG ( K on HG )

    about first comment: step 1 and step 2 are not enough for step 3

    ReplyDelete
  4. AnonymousDecember 9, 2009 at 1:30 PM

    reason for step 1:
    http://i49.tinypic.com/69omdj.gif

    and
    step 1 and step 2 are enough for step 3, i think

    ReplyDelete
  5. Antonio GutierrezDecember 9, 2009 at 2:21 PM

    Thanks!

    ReplyDelete
  6. AjitDecember 9, 2009 at 3:36 PM

    W/o having to refer to any other figure or any construction, we can easily see that quad. OFCE is concyclic and hence ang. HOG = ang. C while ang. OGH = ang. EGC = 90-C/2. Therefore, ang. OHG = 180 – C - (90-C/2)= 90 – C/2 or ang. OHG = ang. OGH. Hence etc.
    Ajit

    ReplyDelete
  7. c .t . e. oDecember 10, 2009 at 1:52 AM

    step 1 and 2 need third condition of congruence,
    angle DOH = angle COG ? verify please

    my solution
    to draw OK perpendicular to HG give
    1) OK is median => HK = KG
    2) OK is diameter perpendicular to chord DC => DK=KC
    so DH=DK-HK
    and GC=KC-KG

    ReplyDelete
  8. Sumith PeirisJune 17, 2016 at 4:53 AM

    Triangles HFC and EGC are similar

    So OH = OG

    So Tr.s ODH and OGC are congruent ASA and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  9. Ignacio Larrosa CañestroOctober 19, 2016 at 9:47 AM

    OGH isosceles ==>
    perpendicular from O to CD bisect GH, and also bisect CD ==>
    GC = DH

    ReplyDelete
  10. Benjamin LeisOctober 21, 2016 at 4:52 PM

    More or less the same as the last few but with a few extra steps included
    1. Right triangle CEG is similar to right triangle CFH because of the angle bisector.
    2. So angle CGE = angle HGO = angle FHC and triangle OHG is isosceles.
    3. OD = OC since they are radii.
    4. So CDO is isoscleses and angle ODC = OCD.
    5. Angle OHD = OGC since they are supplementary to 2 congruent angles.
    6. We now have 2 out of 3 angles and 2 out of 3 sides congruent in ODH and OGC which is more than enough to infer a missing angle and use either SAS or ASA to show the triangles are congruent.
    7. So DH = CG.

    ReplyDelete
  11. MarcoDecember 2, 2023 at 8:25 PM

    Let <DCE=<DCB=x
    <OCE+<OFC=180
    OECF is a cyclic quad
    <HOG=<OCE=2x (ext. < cyclic quad)

    Consider triangle CFH
    <OHG=<FHC=90-x (< sum of triangle)
    <OGH=180-<OHG-<HOG=90-x=<OHG
    So, OH=OG (sides opp. eq. <s)

    Consider triangle OGD & triangle OHC
    OD=OC (radii)
    <ODH=<ODC=<OCD=<OCG (base <s, isos triangle)
    <OGD=<OGH=<OHG=<OHC (Proved)
    Triangle OGD congruent to triangle OHC (AAS)
    So, DG=HC (corr. sides of congruent triangle)
    So, DH+HG=CG+GH
    DH=CG

    ReplyDelete

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