Problem:
In rectangle ABCD,DC=2CB and points E and F lie on AB so that ED and FD trisect ∠ADC as shown. What is the ratio of the area of △DEF to the area of rectangle ABCD?
Answer Choices:
A. 63​​
B. 86​​
C. 1633​​
D. 31​
E. 42​​
Solution:
Let AD=3​. Because ∠ADE=30∘, it follows that AE=1 and DE=2. Now ∠EDF=30∘ and ∠DEF=120∘, so △DEF is isosceles and EF=2. Thus the area of △DEF (with EF viewed as the base) is 21​⋅2⋅3​=3​, and the desired ratio is 3​⋅23​3​​=(A)63​​​.
