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


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

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

  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.

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