Problem:
In △BAC,∠BAC=40∘,AB=10, and AC=6. Points D and E lie on AB and AC, respectively. What is the minimum possible value of BE+DE+CD ?
Answer Choices:
A. 63+3
B. 227
C. 83
D. 14
E. 33+9 Solution:
Let B′ be the reflection of point B across AC, and let C′ be the reflection of point C across AB. Then AB′=AB=10,AC′=AC= 6, BE=B′E,CD=C′D, and ∠B′AC′=120∘. By the Law of Cosines, B′C2=102+62−2⋅10⋅6cos120∘=196; thus B′C′=14. Furthermore, B′C′≤B′E+DE+C′D=BE+DE+CD. Therefore the answer is 14 .