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Geometry ProblemClick the figure below to see the complete problem 502 about Triangle, Two Squares, Midpoint, Perpendicular, Half the measure.
name S midpoint of AC, MB meet AC on P1) draw HT // BD, DT // BH▲BDT = ▲ABC ( S,A,S )=> BT = AC => BM = 1/2 AC2) 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.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.
Problem 502N 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