## Saturday, August 14, 2010

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

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

Complete Problem 502

Level: High School, SAT Prep, College geometry

1. name S midpoint of AC, MB meet AC on P
1) 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°

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

3. Problem 502
N 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