Thursday, August 19, 2010

Problem 506: Triangle with Three Squares, Center, Midpoint, Perpendicular, Congruence

Geometry Problem
Click the figure below to see the complete problem 506 about Triangle with Three Squares, Center, Midpoint, Perpendicular, Congruence.

Problem 506. Triangle with Three Squares, Center, Midpoint, Perpendicular, Congruence


See also:
Complete Problem 506

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 8:17 PM
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2 comments:

  1. Let Q is the projection of E on AC ; N is the projection of M on AC
    Let K is the projection of B on AC; P is the projection of G on AC
    We will prove that N is the midpoint of AC and MN=1/2* AC
    Since M is the midpoint of EQ so N is the midpoint of PQ .
    We have CP=CQ.cos(90-C)= a.sin(C) = BK
    GP=a.cos(C )= CK
    QA= c.cos(A) = BK and EQ=c.cos(A)= AK
    Since CP=QA so N will be midpoint of AC.
    In the trapezoid EQPG we have MN=.5*(EQ+GP)=.5(AK+CK)=.5*AC
    So AMC is a right and isosceles triangle and AMCO is a square.
    Similarly if we project points E, M, B, G over DH and with the same way as above
    We can show that DMH is a right and isosceles triangle .

    Peter Tran

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  2. Aplicamos el teorema de Botema en el segmento AC y obtenemos que AM=MC, que AC=2MX (donde X es proyeccion ortogonal de M en AC) y que <AMC=90.
    Como AO=OC y <AOC=90 tenemos que AOCM es cuadrado.
    by mathreyes

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