Thursday, September 24, 2009

Problem 359. Cyclic quadrilateral, Angle bisector, Rhombus

Proposed Problem
Click the figure below to see the complete problem 359 about cyclic quadrilateral, angle bisector, rhombus.

Problem 359. Cyclic quadrilateral, Angle bisector, Rhombus.
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Complete Problem 359
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 7:59 PM
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2 comments:

  1. Valdir Marques Faria, Colégio VisãoOctober 30, 2009 at 5:32 PM

    Ang(AEN) = ang(DEN) = a, Ang(AFH) = Ang(HFB) = c, ang(ECF) = ang(ECD) = d, ang(EAF) = b, ang(EOF) = x. Assim, teremos que: d = x + a + c , a + c = d – x (1) e x = b + a + c , a + c = x – b (2). De (1) e (2), temos que: x – b = d – x , x = d + b = 90°, pois ABCD é um quadrilátero cíclico. Assim, MN  HG. Como EO perpendicular a HG e EO é bissetriz do ângulo AED, temos que o triângulo EHG é isósceles, concluindo que EO é também mediana do triângulo EHG em relação ao lado HG. Assim, HG = OG. De modo análogo, OM = ON. Como O é ponto médio das diagonais MN e HG, HMGN é um paralelogramo. Como as diagonais HG e MN são perpendiculares, então o paralelogramo é um losango.

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  2. Peter TranFebruary 3, 2017 at 7:56 PM

    https://goo.gl/photos/zmDGSDPNEzNTZ8jA7

    Define angles a, b, d, e, f , u and point P as per sketch
    Locate point Q on AD such that ∠FCD=∠DCQ= u
    1. triangles CDF and ABF are similar (case AA) => FD/FC=FB/FA
    Since FH bisect angle AFB => GD/GC= FD/FC=FB/FA=HB/HA …. (1)
    Similarly we also have MB/MC= ND/NA…. (2)
    Triangles CBE and CDQ are similar ( case AA)=> EB/EC=QD/QC=MB/MC=ND/NA…. (3)
    Since CD bisect angle QCF => QD/QC=FD/FC=GD/GC= HB/HA…. (4)
    Compare (1), (2), (3) and (4) => MB/MC= GD/GC=HB/HA=ND/NA
    So MG//BD//HN and MG=HN => quadrilateral HMGN is a parallelogram
    2. in the complete quadrilateral EHANFP we will have ∠EPF= a+ 1/2e +1/2f…. (5)
    In the complete quadrilateral EMPGFC we will have ∠ECF= ^EPF+ 1/2e +1/2f… (6)
    replace (5) to (6) and note that ∠ECF supplement to angle a , we will get ∠EPF= 90
    parallelogram HMGN become a rhombus




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