Saturday, September 12, 2009

Problem 354. Rhombus, Square, 45 degrees

Proposed Problem
Click the figure below to see the complete problem 354 about Rhombus, Square, 45 degrees.

 Problem 354. Rhombus, Square, 45 degrees.
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Complete Problem 354
Level: High School, SAT Prep, College geometry

8 comments:

  1. Seja H o ponto de intersecção de CB e DE. O Ang(AGD) = Ang(CGH) = x. Assim, Ang(CDE) = Ang(DEC) = a, pois CE = CD. Ang(CHE) = 90° - a, pois o triângulo CEH é retângulo em C. O ang(DCA) = Ang(ACB) = b, pois a diagonal AC do losango (rombo) é também uma bissetriz do ângulo C. No triângulo CGH, temos que 90° - a = x + b, ou seja, x = 90° - (a + b), pelo teorema do ângulo externo. No triângulo CDG, temos que x = a + b, pelo mesmo teorema. Assim, x = 90° - x e, concluindo, temos que x = 45°(demonstrado).

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  2. join B with G and D.

    1) BG = GD (G equal distance to B,D)
    2) ang CBD = ang CDB ( BC=DC )

    => angCBG = angCDG (3)

    4) ang CFG = ang CDG ( FC = DC )

    from 3 and 4

    ang CBG = ang CFG

    but CF perpendicular to CB, so BG must be perpend to FG
    => BGD = 90
    => AGD = 45 ( CA bisector of C )

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    Replies
    1. But how is cf perp to cb?? Whats your reasoning?

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  3. Triangle ECD is isosceles.
    Let angleACD = x = angleBCA , angle CED = y = angle CDE.
    By considering triangleECD,
    angleDEC + angleECD + angleCDE = 180
    => (y) + (90 + x + x) + (y) = 180
    => x + y = 90
    => angleGCD + angleCDG = 90
    => angleAGD = 90

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  4. http://s22.postimg.org/8ux75gpb5/pro_354.png
    Connect AF and FC
    Observe that AF//DE
    And BF=BA=BC => B is the center of circumcircle of triangle AFC
    Inscribed angle FAC= ½ of central angle FBC= angle AGD=45

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  5. Construct the square DCMN outside the rhombus; triangles CED and CBM are congruent and have the homologous sides perpendicular, consequently DE_|_BM and they intersect on AC due to symmetry, thus AC is the bisector of <EFM=90.

    Best regards

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  6. Problem 354
    Let K,L are centers of BCEF , ABCD respectively, then <BKC=90=<BLC so BLCK is cyclic with
    <KLC=<KBC=45. But BL=LD, BK=KE then LK//DE .So <AGD=<KLC=45.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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  7. CB = CE = CD, hence C is the circumcenter of triangle BDE.

    So < BED = < BCD/2 = < BCG, thus BGCE is concyclic.

    Therefore < AGD = < CGE = < CBE = 45.

    Sumith Peiris
    Moratuwa
    Sri Lanka

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