Problem:
Point E is the midpoint of side CD in square ABCD, and BE meets diagonal AC at F. The area of quadrilateral AFED is 45. What is the area of ABCD?
Answer Choices:
A. 100
B. 108
C. 120
D. 135
E. 144
Solution:
Let H be the point on BC where the altitude from F to BC meets BC. This altitude, FH, is illustrated above. Then, by angle-angle similarity, we can see that △CAB∼△CFH and △BFH∼△BEC. Since the sides of similar triangles are proportional, we know that BHFH​=BCEC​ and HCFH​=BCAB​. Thus, ECFH​=BCBH​ and ABFH​=BCHC​.
Adding these equations yields
ECFH​+ABFH​=BCBH​+BCHC​=BCBH+HC​=BCBC​=1.
This, in turn, goes to show that EC1​+AB1​=FH1​.
Now, let s be the side length of the square. We know AB=2⋅EC=s. This means FH1​=EC1​+AB1​=a1​+a2​=a3​. Therefore, FH=3s​.
Now, to compute the area of â–³EFC, we take the area of â–³BCE and subtract the area of â–³BFC. This is equal to
BC⋅EC2−BC⋅FH2=BC⋅(EC−FH)2=s⋅(2s​−3s​)2=s⋅(6s​)2=12s2​.
The area of AFED is the area of â–³ACD minus the area of â–³EFC, which is equal to
2s2​−12s2​=125s2​=45.
With 125​s2=45, we get s2=108, which is the area of the full square.
Answer: B​.
The problems on this page are the property of the MAA's American Mathematics Competitions
