Problem:
Quadrilateral ABCD has right angles at B and C,△ABC∼△BCD, and AB>BC. There is a point E in the interior of ABCD such that △ABC∼△CEB and the area of △AED is 17 times the area of △CEB. What is BCAB ?
Answer Choices:
A. 1+2
B. 2+2
C. 17
D. 2+5
E. 1+23 Solution:
Let F lie on AB so that DF⊥AB. Because BCDF is a rectangle, ∠FCB≅∠DBC≅∠CAB≅∠BCE, so E lies on CF and it is the foot of the altitude to the hypotenuse in △CBF. Therefore △BEF∼△CBF≅△BCD∼△ABC. Because
DF⊥AB,FE⊥EB, and DFAB=BCAB=FEBE
it follows that △ABE∼△DFE. Thus ∠DEA=∠DEF−∠AEF=∠AEB−∠AEF=∠FEB=90∘. Furthermore,
EDAE=EFBE=BCAB
so △AED∼△ABC. Assume without loss of generality that BC=1, and let AB=r>1. Because BCAB=CDBC, it follows that BF=CD=r1. Then
17=Area(△CEB)Area(△AED)=AD2=FD2+AF2=1+(r−r1)2,
and because r>1 this yields r2−4r−1=0, with positive solution r=2+5.
OR
Without loss of generality, assume that BC=1. The given conditions imply that the quadrilateral can be placed in the coordinate plane with C=(0,0),B=(0,1),A=(r,1), and D=(r1,0). Let E have positive coordinates (x,y). Because △ABC∼△CEB, these coordinates must satisfy
yx=tan(∠ECB)=tan(∠BAC)=r1
and
x2+y2=1CE=1+r2r
Solving this system of equations gives
x=1+r2r and y=1+r2r2
The area of △CEB is 2x. The area of △AED can be computed using the fact that the area of a polygon with vertices (x1,y1),(x2,y2),…, (xn,yn) in counterclockwise order is
Substituting in the expressions for x and y in terms of r, setting Area(△AED)=17⋅Area(△CEB), and simplifying yields the equation r4−18r2+1=0. Applying the quadratic formula, and noting that r>1, gives r2=9+45=(2+5)2, so r=2+5.
OR
Let θ=∠ACB, and without loss of generality assume BC=1. Let F lie on AB so that DF⊥AB. Then the requested fraction is AB=tanθ. Because △ABC∼△BCD∼△CEB∼△BEF, it follows that CD=cotθ,BE=cosθ, and CE=sinθ. Then the area of quadrilateral ABCD is [ABCD]=21(tanθ+cotθ)=2sinθcosθ1; and the areas of three of the four triangles into which that area can
be decomposed are [ABE]=21tanθcos2θ=21sinθcosθ,[BCE]=21sinθcosθ, and [CDE]=21sin2θcotθ=21sinθcosθ. (Interestingly, the three triangles all have the same area.) Then
[AED]=2sinθcosθ1−23sinθcosθ=17⋅21sinθcosθ
This last equation simplifies to 20sin2θcos2θ=1, so (2sinθcosθ)2=51. Then sin(2θ)=51,cos(2θ)=5−2 (because AB>BC implies 4π<θ<2π), and
