Problem:
If ABCD is a 2×2 square, E is the midpoint of AB,F is the midpoint of BC,AF and DE intersect at I, and BD and AF intersect at H, then the area of quadrilateral BEIH is
Answer Choices:
A. 31
B. 52
C. 157
D. 158
E. 53
Solution:
Use coordinates with B=(0,0),F=(1,0) and E=(0,1). The equations of lines BH,IF and EI are y=x,y=−2x+2 and y=21x+1, respectively. Thus H=(32,32) and I=(52,56). Hence the altitude of △BHF from vertex H is 2/3, and thus [BHF]=1/3.† Similarly, the altitude of △AIE from vertex I is 2/5, so [AIE]=1/5. Therefore
[BEIH]=[BAF]−[AIE]−[BHF]=1−51−31=157
OR
Triangles DAE and ABF have equal sides so they are congruent. Thus ∠EAI+∠AEI=∠EAI+∠BFA=90∘, and △AIE is a right triangle similar to △ABF. Since [ABF]=1 and AF=12+22=5, we have
[AIE]=[ABF][AIE]=AF2AE2=51.
Note that △BHF is similar to △DHA because of equal angles, and that the ratio of similarity is DABF=21. Hence HAHF=21 and AFHF=31. Thus, since △BHF and △BAF share side BF and the altitudes to that side are in the ratio 1:3,[BHF]=[BAF][BHF]=31.
Hence [BEIH]=[BAF]−[AIE]−[BHF]=1−51−31=157.
OR
Let the areas of triangles AEI,EHI,BHE and BHF be w,x,y and z, respectively. Since BE=BF and ∠EBH=∠FBH, triangles BHE and BHF are congruent. Hence, y=z.
Since EH is a median of △AHB, we have w+x=y. Hence
3y=(w+x)+y+z=21BF⋅AB=1, or y=31.
If △ABF is rotated 90∘ clockwise about the center of the square, it coincides with △DAE. Hence AF⊥DE, from which it follows that △AIE∼△DAE. Hence DAAI=AEIE=DEAE=51, and w=(51)2[DAE]=51. Since w+x=y, we have [BEIH]=x+y=2y−w=32−51=157.
