Geometry Problem

Click the figure below to see the complete problem 502 about Triangle, Two Squares, Midpoint, Perpendicular, Half the measure.

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Complete Problem 502

Level: High School, SAT Prep, College geometry

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## Saturday, August 14, 2010

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Problem 502: Triangle, Two Squares, Midpoint, Perpendicular, Half the measure

See also:

Complete Problem 502

Level: High School, SAT Prep, College geometry

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Geometry Problem

Click the figure below to see the complete problem 502 about Triangle, Two Squares, Midpoint, Perpendicular, Half the measure.

See also:

Complete Problem 502

Level: High School, SAT Prep, College geometry

Labels:
midpoint,
perpendicular,
square,
triangle

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name S midpoint of AC, MB meet AC on P

ReplyDelete1) draw HT // BD, DT // BH

▲BDT = ▲ABC ( S,A,S )

=> BT = AC => BM = 1/2 AC

2) MBP = A + 90 + ABP = 180 => ABP = 90 - A =>

=> APB = 90°

Let P be midpoint of AC, and R be foot of altitude from B to AC.

ReplyDeleteLet centers of squares EDBA and GHBC be C1 and C2.

From previous problem, PC1MC2 is square, so that <C1PC2=90.

Based on cyclic quadrilaterals ARBC1 and CRBC2, <C1RA=<C1BA=45 and <C2RC=<C2BC=45, so that <C1RC2=90 as well.

This puts P, C1, M, C2, and R on same circle. It follows that <PRM=180-<MC1P=90=<PRB (last equality by definition), making M, B, and R collinear.

For other half of stated problem, triangle BC1M is rotation of triangle AC1P about C1.

Problem 502

ReplyDeleteN takes the symmetry of B with respect to M, then the DNHB is parallelogram.Then DN=BH=BC, DB=BA and <ABC=180-<DBH=<NDB so triangle DBN=triangle ABC.

Therefore BN=AC or BM=AC/2 and <DBN=<BAK(K=sectional BN and AC) but

<DBN+<ABK=90 or <BAK+<ABK=90. Therefore MB is perpendicular to AC.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE