Geometry Problem

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

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Complete Problem 497

Level: High School, SAT Prep, College geometry

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## Tuesday, August 10, 2010

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

See also:

Complete Problem 497

Level: High School, SAT Prep, College geometry

## Link List

Online Geometry theorems, problems, solutions, and related topics.

Geometry Problem

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

See also:

Complete Problem 497

Level: High School, SAT Prep, College geometry

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ON midline of ▲AGC

ReplyDeleteOM midlime of ▲ADC

AG = DC , AG perpendicular DC

=>

OM = ON, OM perpendicular to ON

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

ReplyDelete1. Add the other midpoints in for symmetry: X on AB and Y on BC. This forms

ReplyDeletea 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.

∆ABG≡∆DBC (AB=DB, BG=BC), ∡ABG = ∡CBD) => AG = DC => ON = OM

ReplyDeleteFrom congruence, ∡CDB = ∡AGB ( perp sides ) => AG ⊥ DC, => ON ⊥ OM