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.
See more:
Complete Problem 354
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 11:40 AM
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11 comments:

  1. Valdir Marques Faria, Colégio VisãoOctober 27, 2009 at 12:19 PM

    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).

    ReplyDelete
  2. c .t . e. oDecember 18, 2009 at 7:31 AM

    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 )

    ReplyDelete
    Replies
    1. UnknownMarch 12, 2013 at 5:11 PM

      But how is cf perp to cb?? Whats your reasoning?

      Delete
    2. c.t.e.o.December 15, 2020 at 10:51 AM

      A lot of mistakes sure by changing point names
      https://photos.app.goo.gl/T9M1f86BJQ4cGcTY6

      Delete
    3. c.t.e.o.December 15, 2020 at 11:03 AM

      https://photos.app.goo.gl/ogSWzeoZid1UPQ3B8

      Delete
  3. W FungMarch 13, 2013 at 9:19 AM

    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

    ReplyDelete
  4. Peter TranDecember 25, 2015 at 11:03 PM

    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

    ReplyDelete
  5. Stan FulgerDecember 27, 2015 at 11:32 PM

    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

    ReplyDelete
  6. APOSTOLIS MANOLOUDISAugust 12, 2016 at 2:11 PM

    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

    ReplyDelete
  7. Sumith PeirisOctober 18, 2016 at 3:48 AM

    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

    ReplyDelete
  8. rv.littlemanDecember 13, 2020 at 7:11 AM

    See the drawing

    ABCD is a rhombus => ΔACD is congruent to ΔACB
    => ∠ACD = ∠ACB
    => ΔGCD is congruent to ΔGCB (SAS)
    => ∠GDC = ∠GBC
    ABCD is a rhombus =>CE=CD => ΔDCE is isosceles in C
    => ∠GEC = ∠GDC
    ∠GEC = ∠GBC => B, G, C, E are concyclic

    ∠AGD = ∠EGC
    Arc CE : ∠EBC = ∠EGC
    ∠EBC = 45° therefore ∠EGC = 45°

    ReplyDelete

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