## 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.

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

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).

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 )

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

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

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

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

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

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