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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## 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

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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

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.

ReplyDeleteAlso 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

See the

ReplyDeletedrawing∠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

DeleteTherefore ∠MON=π/2