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

5 comments:

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

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

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  3. 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.

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

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  5. 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

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