Tuesday, August 10, 2010

Problem 497: Triangle, Two Squares, Centers, Midpoint, 90 Degrees

Geometry Problem
Click the figure below to see the complete problem 497 about Triangle, Two Squares, Centers, Midpoint, 90 Degrees, Congruence.

Problem 497: Triangle, Two Squares, Centers, Midpoint, 90 Degrees.


See also:
Complete Problem 497

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 2:04 PM
Labels: , , , , , ,

7 comments:

  1. c .t . e. oAugust 10, 2010 at 2:13 PM

    ON midline of ▲AGC
    OM midlime of ▲ADC
    AG = DC , AG perpendicular DC
    =>
    OM = ON, OM perpendicular to ON

    ReplyDelete
  2. UnknownApril 21, 2017 at 1:43 AM

    I see AG=DC. I don't understand that AG perpensicular DC. Excuse me please explain.

    ReplyDelete
  3. Benjamin LeisApril 21, 2017 at 11:49 AM

    1. Add the other midpoints in for symmetry: X on AB and Y on BC. This forms
    a parallelogram XBYO and 3 similar triangles ABC, COY and AOX.
    2. tr MOX is congruent to tr NOY by SAS. (Note: they are inverted versions) OX = BY from the parallelogram. OY = BX for the same reason. NY = BY since BNY is a right isosceles. Similarly MX = BX. The middle angles (MXO and NYC are congruent from an angle chase: 90 + angle B in both cases.)
    3. This gives you that OM = ON.
    4.That also shows that angle XOM and YON are two of the three angles in the
    congruent triangles. So XOM + YON = 180 - (OXM or OYN). Each of these angles is 90 + angle B. And from the parallegoram angle XOY is also angle B.
    So putting those together angle MON = 180 - (90 + angle B) + angle B = 90.

    ReplyDelete
  4. c.t.e.oApril 22, 2017 at 10:33 AM

    ∆ABG≡∆DBC (AB=DB, BG=BC), ∡ABG = ∡CBD) => AG = DC => ON = OM
    From congruence, ∡CDB = ∡AGB ( perp sides ) => AG ⊥ DC, => ON ⊥ OM

    ReplyDelete
  5. Sumith PeirisNovember 18, 2017 at 2:43 AM

    From my solution to problem 496, AG = CD and so OM = DC/2 =AG/2 = O using the mid point theorem on Tr.s ADC and AGC.

    Also from my solution to Problem 496,CD is perpendicular to AG and so
    < GAC + < ACD = 90.
    Since < GAC = < NOC and < ACD = < MOA,
    < NOC + < MOA = 90 and so < MON = 90

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  6. rv.littlemanDecember 13, 2021 at 11:10 AM

    See the drawing

    ∠DBC = ∠DBA+∠ABC = π/2+∠ABC
    ∠ABG = ∠ABC+∠BCG = ∠ABC+ π/2
    => ∠DBC = ∠ABG
    => Δ DBC is congruent to Δ ABG (SAS)
    => ∠BDC = ∠BAG and DC=AG

    Define K the intersection of AG and BC
    => ∠BDC = ∠BAG => ∠BDK = ∠BAK => AKBDE are concyclic
    => ∠AKD = ∠ABD = π/2

    In the ΔACG :O in the middle of AC and N in the middle of CG
    => ON//AG and ON=AG/2
    In the ΔACD :O in the middle of AC and M in the middle of AD
    => OM//CD and OM=CD/4
    Therefore OM=ON

    ∠AKD = ∠ABD = π/2 and ON//AG and OM//CD
    Therefore ∠MON = ��/2

    ReplyDelete

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